Rs Aggarwal 2019 for Class 10 Math Chapter 3 - Linear Equations In Two Variables

Share with your friends

Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 3 Linear Equations In Two Variables are provided here with simple step-by-step explanations. These solutions for Linear Equations In Two Variables are extremely popular among Class 10 students for Math Linear Equations In Two Variables Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 87:

Question 1:

$2 x + 3 y = 2, x - 2 y = 8 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
Graph of 2x + 3y = 2

2x + 3y = 2
⇒ 3y = (2 − 2x)
⇒ 3y = 2(1 − x)
⇒ y = $\frac{2 (1 - x)}{3}$ ...(i)
Putting x = 1, we get y = 0
Putting x = − 2, we get y = 2
Putting x = 4, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 2.

x	1	−2	4
y	0	2	−2

Now, plot the points A(1, 0), B(− 2 , 2) and C(4, − 2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, the line BC is the graph of 2x + 3y = 2.

Graph of x − 2y = 8
x − 2y = 8
⇒ 2y = ( x − 8)
⇒

y = \frac{x - 8}{2}

...(ii)
Putting x = 2, we get y = −3
Putting x = 4, we get y = −2
Putting x = 0, we get y = −4
Thus, we have the following table for the equation x - 2y = 8.

x	2	4	0
y	− 3	− 2	− 4

Now, plot the points P(0, − 4) and Q(2, − 3). The point C(4, −2) has already been plotted. Join PQ and QC and extend it on both ways.
Thus, line PC is the graph of x − 2y = 8.

The two graph lines intersect at C(4, −2).
∴ x = 4 and y = −2 are the solutions of the given system of equations.

Page No 87:

Question 2:

$3 x + 2 y = 4, 2 x - 3 y = 7 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' representing the x-axis and y-axis, respectively.
Graph of 3x + 2y = 4

3x + 2y = 4
⇒ 2y = (4 − 3x)
⇒ y = $\frac{4 - 3 x}{2}$ ...(i)
Putting x = 0, we get y = 2
Putting x = 2, we get y = −1
Putting x = −2, we get y = 5
Thus, we have the following table for the equation 3x + 2y = 4.

x	0	2	− 2
y	2	− 1	5

Now, plot the points A(0, 2), B( 2, −1) and C(−2, 5) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + 2y = 4.

Graph of 2x − 3y = 7
2x − 3y = 7
⇒ 3y = (2x − 7)
⇒

y = \frac{2 x - 7}{3}

Putting x = 2, we get y = −1
Putting x = −1, we get y = −3
Putting x = 5, we get y = 1
Thus, we have the following table for the equation 2x − 3y = 7.

x	2	−1	5
y	−1	−3	1

Now, plot the points P(−1, −3) and Q(5, 1). The point B(2, −1) has already been plotted. Join PB and QB and extend it on both ways.
Thus, PQ is the graph of 2x − 3y = 7.

The two graph lines intersect at B(2, − 1).
∴ x = 2 and y = −1 are the solutions of the given system of equations.

Page No 87:

Question 3:

$2 x + 3 y = 8, x - 2 y + 3 = 0 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 8

2x + 3y = 8
⇒ 3y = (8 − 2x)
∴ $y = \frac{8 - 2 x}{3}$ ...........(i)
Putting x = 1, we get y = 2.
Putting x = −5, we get y = 6.
Putting x = 7, we get y = −2.
Thus, we have the following table for the equation 2x + 3y = 8.

x	1	−5	7
y	2	6	−2

Now, plot the points A(1, 2), B(5, −6) and C(7, −2) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 2x + 3y = 8.

Graph of x − 2y + 3 = 0
x − 2y + 3 = 0
⇒ 2y = (x + 3)
∴

y = \frac{x + 3}{2}

..........(ii)
Putting x = 1, we get y = 2.
Putting x = 3, we get y = 3.
Putting x = −3, we get y = 0.
Thus, we have the following table for the equation x − 2y + 3 = 0.

x	1	3	−3
y	2	3	0

Now, plot the points P(3, 3) and Q(−3, 0). The point A(1, 2) has already been plotted. Join AP and QA and extend it on both ways.
Thus, PQ is the graph of x − 2y + 3 = 0.

The two graph lines intersect at A(1, 2).
∴ x = 1 and y = 2

Page No 87:

Question 4:

$2 x - 5 y + 4 = 0, 2 x + y - 8 = 0 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x − 5y + 4 = 0

2x − 5y + 4 = 0
⇒ 5y = (2x + 4)
∴ $y = \frac{2 x + 4}{5}$ ...........(i)
Putting x = −2, we get y = 0.
Putting x = 3, we get y = 2.
Putting x = 8, we get y = 4.
Thus, we have the following table for the equation 2x − 5y + 4 = 0.

x	−2	3	8
y	0	2	4

Now, plot the points A(−2 , 0), B(3, 2) and C(8, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x − 5y + 4 = 0.

Graph of 2x + y − 8 = 0
2x + y − 8 = 0
⇒ y = (8 − 2x) ..........(ii)
Putting x = 1, we get y = 6.
Putting x = 3, we get y = 2.
Putting x = 2, we get y = 4.
Thus, we have the following table for the equation 2x + y − 8 = 0.

x	1	3	2
y	6	2	4

Now, plot the points P(1, 6) and Q(2, 4). The point B(3, 2) has already been plotted. Join PQ and QB and extend it on both ways.
Thus, PB is the graph of 2x + y − 8 = 0.

The two graph lines intersect at B(3, 2).
∴ x = 3 and y = 2

Page No 87:

Question 5:

Solve the following system of equations graphically:

3 x + 2 y = 12 5 x - 2 y = 4

Answer:

The given equations are:
$3 x + 2 y = 12 . . . . . (i) 5 x - 2 y = 4 . . . . . (i i)$
From (i), write y in terms of x
$y = \frac{12 - 3 x}{2} . . . . . (i i i)$
Now, substitute different values of x in (iii) to get different values of y
For x = 0, $y = \frac{12 - 3 x}{2} = \frac{12 - 0}{2} = 6$
For x = 2, $y = \frac{12 - 3 x}{2} = \frac{12 - 6}{2} = 3$
For x = 4, $y = \frac{12 - 3 x}{2} = \frac{12 - 12}{2} = 0$
Thus, the table for the first equation (3x + 2y = 12) is

x	0	2	4
y	6	3	0

Now, plot the points A(0, 6), B(2, 3) and C(4, 0) on a graph paper and join
A, B and C to get the graph of 3x + 2y = 12.
From (ii), write y in terms of x

y = \frac{5 x - 4}{2} . . . . . (i v)

Now, substitute different values of x in (iv) to get different values of y
For x = 0,

y = \frac{5 x - 4}{2} = \frac{0 - 4}{2} = - 2

For x = 2,

y = \frac{5 x - 4}{2} = \frac{10 - 4}{2} = 3

For x = 4,

y = \frac{5 x - 4}{2} = \frac{20 - 4}{2} = 8

Thus, the table for the first equation (5x − 2y = 4) is

x	0	2	4
y	−2	3	8

Now, plot the points D(0, −2), E(2, 3) and F(4, 8) on the same graph paper and join
D, E and F to get the graph of 5x − 2y = 4.

From the graph it is clear that, the given lines intersect at (2, 3).
Hence, the solution of the given system of equations is (2, 3).

Page No 87:

Question 6:

$3 x + y + 1 = 0, 2 x - 3 y + 8 = 0 .$

Answer:

n a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x - axis and y - axis respectively.
Graph of 3x + y + 1 = 0

3x + y + 1 = 0
⇒ y = (−3x − 1) ...........(i)
Putting x = 0, we get y = −1.
Putting x = −1, we get y = 2.
Putting x = 1, we get y = −4.
Thus, we have the following table for the equation 3x + y + 1 = 0.

x	0	−1	1
y	−1	2	−4

Now, plot the points A(0, −1), B(−1, 2) and C(1, −4) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x + y + 1 = 0.

Graph of 2x − 3y + 8 = 0
2x − 3y + 8 = 0
⇒ 3y = (2x + 8)
∴

y = \frac{2 x + 8}{3}

Putting x = −1, we get y = 2.
Putting x = 2, we get y = 4.
Putting x = −4, we get y = 0.
Thus, we have the following table for the equation 2x − 3y + 8 = 0.

x	−1	2	−4
y	2	4	0

Now, plot the points P(2, 4), Q(−4, 0). The point B(−1 , 2) has already been plotted. Join PB and BQ and extend it on both ways.
Thus, PQ is the graph of 2x − 3y + 8 = 0.

The two graph lines intersect at B(−1, 2).
∴ x = −1 and y = 2

Page No 87:

Question 7:

Solve the following system of equation graphically:
$2 x + 3 y + 5 = 0 3 x - 2 y - 12 = 0$

Answer:

From the first equation, write y in terms of x
$y = - (\frac{5 + 2 x}{3}) . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 1, y = - \frac{5 - 2}{3} = - 1 For x = 2, y = - \frac{5 + 4}{3} = - 3 For x = 5, y = - \frac{5 + 10}{3} = - 5$
Thus, the table for the first equation ( 2x + 3y + 5 = 0 ) is

x	−1	2	5
y	−1	−3	−5

Now, plot the points A(−1,−1), B(2,−3) and C(5,−5) on a graph paper and join
them to get the graph of 2x + 3y + 5 = 0.
From the second equation, write y in terms of x

y = \frac{3 x - 12}{2} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \frac{0 - 12}{2} = - 6 For x = 2, y = \frac{6 - 12}{2} = - 3 For x = 4, y = \frac{12 - 12}{2} = 0

So, the table for the second equation ( 3x − 2y − 12 = 0 ) is

x	0	2	4
y	−6	−3	0

Now, plot the points D(0,−6), E(2,−3) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 3x − 2y − 12 = 0.

From the graph it is clear that, the given lines intersect at (2,−3).
Hence, the solution of the given system of equation is (2,−3).

Page No 87:

Question 8:

Solve the following system of equation graphically:
$2 x - 3 y + 13 = 0 3 x - 2 y + 12 = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{2 x + 13}{3} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 5, y = \frac{- 10 + 13}{3} = 1 For x = 1, y = \frac{2 + 13}{3} = 5 For x = 4, y = \frac{8 + 13}{3} = 7$
Thus, the table for the first equation ( 2x − 3y + 13 = 0 ) is

x	−5	1	4
y	1	5	7

Now, plot the points A(−5,1), B(1,5) and C(4,7) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 13 = 0.
From the second equation, write y in terms of x

y = \frac{3 x + 12}{2} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 4, y = \frac{- 12 + 12}{2} = 0 For x = - 2, y = \frac{- 6 + 12}{2} = 3 For x = 0, y = \frac{0 + 12}{2} = 6

So, the table for the second equation ( 3x − 2y + 12 = 0 ) is

x	−4	−2	0
y	0	3	6

Now, plot the points D(−4,0), E(−2,3) and F(0,6) on the same graph paper and join
D, E and F to get the graph of 3x − 2y + 12 = 0.

From the graph it is clear that, the given lines intersect at (−2,3).
Hence, the solution of the given system of equation is (−2,3).

Page No 87:

Question 9:

$2 x + 3 y = 4, 3 x - y = - 5 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4

2x + 3y = 4
⇒ 3y = (4 − 2x)
∴ $y = \frac{4 - 2 x}{3}$ ...(i)
Putting x = −1, we get y = 2
Putting x = 2, we get y = 0
Putting x = 5, we get y = −2
Thus, we have the following table for the equation 2x + 3y = 4.

x	−1	2	5
y	2	0	−2

Now, plot the points A(−1 , 2), B(2 , 0) and C(5, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.

                                                 Graph of 3x − y = −5
3x − y = −5
⇒ y = (3x + 5)              ...(ii)
Putting x = −1, we get y = 2
Putting x = 0, we get y = 5
Putting x = −2, we get y = −1
Thus, we have the following table for the equation 3x − y = − 5 = 0.

x	−1	0	−2
y	2	5	−1

Now, plot the points P(0, 5), Q(−2 , −1). The point A(−1 , 2) has already been plotted. Join PA and QA and extend it on both ways.
Thus, PQ is the graph of 3x − y = −5.

The two graph lines intersect at A(−1 , 2).
∴ x = −1 and y = 2 are the solutions of the given system of equations.

Page No 87:

Question 10:

$x + 2 y + 2 = 0, 3 x + 2 y - 2 = 0 .$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x + 2y + 2 = 0

x + 2y + 2 = 0
⇒ 2y = (−2 − x)
∴ $y = \frac{- 2 - x}{2}$ ...............(i)
Putting x = −2, we get y = 0.
Putting x = 0, we get y = −1.
Putting x = 2, we get y = −2.
Thus, we have the following table for the equation x + 2y + 2 = 0.

x	−2	0	2
y	0	−1	−2

Now, plot the points A(−2, 0) , B(0 , −1) and C(2, −2) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x + 2y + 2 = 0.

Graph of 3x + 2y − 2 = 0
3x + 2y − 2 = 0
⇒ 2y = (2 − 3x)
∴

y = \frac{2 - 3 x}{2}

...............(ii)
Putting x = 0, we get y = 1.
Putting x = 2, we get y = −2.
Putting x = 4, we get y = −5.
Thus, we have the following table for the equation 3x + 2y − 2 = 0.

x	0	2	4
y	1	−2	−5

Now, plot the points P(0, 1) and Q(4, −5). The point C(2, −2) has already been plotted. Join PC and QC and extend it on both ways.
Thus, PQ is the graph of 3x + 2y − 2 = 0.

The two graph lines intersect at A(2, −2).
∴ x = 2 and y = −2

Page No 87:

Question 11:

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the x-axis:
$x - y + 3 = 0 2 x + 3 y - 4 = 0$

Answer:

From the first equation, write y in terms of x
$y = x + 3 . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 3, y = - 3 + 3 = 0 For x = - 1, y = - 1 + 3 = 2 For x = 1, y = 1 + 3 = 4$
Thus, the table for the first equation (x − y + 3 = 0) is

x	−3	−1	1
y	0	2	4

Now, plot the points A(−3,0), B(−1,2) and C(1,4) on a graph paper and join
A, B and C to get the graph of x − y + 3 = 0.
From the second equation, write y in terms of x

y = \frac{4 - 2 x}{3} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 4, y = \frac{4 + 8}{3} = 4 For x = - 1, y = \frac{4 + 2}{3} = 2 For x = 2, y = \frac{4 - 4}{3} = 0

So, the table for the second equation ( 2x + 3y − 4 = 0 ) is

x	−4	−1	2
y	4	2	0

Now, plot the points D(−4,4), E(−1,2) and F(2,0) on the same graph paper and join
D, E and F to get the graph of 2x − 3y − 4 = 0.

From the graph it is clear that, the given lines intersect at (−1,2).
So, the solution of the given system of equation is (−1,2).
The vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2) and (2,0).
Now, draw a perpendicular from the intersection point E on the x-axis. So,

Area (∆ E A F) = \frac{1}{2} \times A F \times E M = \frac{1}{2} \times 5 \times 2 = 5 sq . units

Hence, the vertices of the triangle formed by the given lines and the x-axis are (−3,0), (−1,2)
and (2,0) and its area is 5 sq. units.

Page No 87:

Question 12:

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the x-axis:
$2 x - 3 y + 4 = 0 x + 2 y - 5 = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{2 x + 4}{3} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 2, y = \frac{- 4 + 4}{3} = 0 For x = 1, y = \frac{2 + 4}{3} = 2 For x = 4, y = \frac{8 + 4}{3} = 4$
Thus, the table for the first equation (2x − 3y + 4 = 0) is

x	−2	1	4
y	0	2	4

Now, plot the points A(−2,0), B(1,2) and C(4,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 4 = 0.
From the second equation, write y in terms of x

y = \frac{5 - x}{2} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 3, y = \frac{5 + 3}{2} = 4 For x = 1, y = \frac{5 - 1}{2} = 2 For x = 5, y = \frac{5 - 5}{2} = 0

So, the table for the second equation ( x + 2y − 5 = 0 ) is

x	−3	1	5
y	4	2	0

Now, plot the points D(−3,4), B(1,2) and F(5,0) on the same graph paper and join
D, E and F to get the graph of x + 2y − 5 = 0.

From the graph it is clear that, the given lines intersect at (1,2).
So, the solution of the given system of equations is (1,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (1,2) and (5,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,

Area (∆ B A F) = \frac{1}{2} \times A F \times B M = \frac{1}{2} \times 7 \times 2 = 7 sq . units

Hence, the vertices of the triangle formed by the given lines and the x-axis are (−2,0), (1,2) and (5,0)
and the area of the triangle is 7 sq. units.

Page No 87:

Question 13:

Solve the following system of linear equations graphically:
$4 x - 3 y + 4 = 0, 4 x + 3 y - 20 = 0 .$
Find the area of the region bounded by these lines and the x-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 4x − 3y + 4 = 0
4x − 3y + 4 = 0
⇒ 3y = (4x + 4)
∴ $y = \frac{4 x + 4}{3}$ ............(i)
Putting x = −1, we get y = 0.
Putting x = 2, we get y = 4.
Putting x = 5, we get y = 8.
Thus, we have the following table for the equation 4x − 3y + 4 = 0

x	−1	2	5
y	0	4	8

Now, plot the points A(−1, 0), B( 2, 4) and C(5, 8) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 4x − 3y + 4 = 0.

Graph of 4x + 3y − 20 = 0
4x + 3y − 20 = 0
⇒ 3y = (−4x + 20)
∴

y = \frac{- 4 x + 20}{3}

............(ii)
Putting x = 2, we get y = 4.
Putting x = −1, we get y = 8.
Putting x = 5, we get y = 0.
Thus, we have the following table for the equation 4x + 3y − 20 = 0.

x	2	−1	5
y	4	8	0

Now, plot the points P(1, −8) and Q(5 , 0). The point B(2, 4) has already been plotted. Join PB and QB to get the graph line PQ. Extend it on both ways.
Then, line PQ is the graph of the equation 4x + 3y − 20 = 0.

The two graph lines intersect at B(2, 4).
∴ The solution of the given system of equations is x = 2 and y = 4.
Clearly, the vertices of ΔABQ formed by these two lines and x-axis are Q(5, 0), B(2, 4) and A(−1, 0).
Now, consider ΔABQ.
Here, height = 4 units and base (AQ) = 6 units
∴ Area of ΔABQ =

\frac{1}{2} \times base \times height

sq. units
=

(\frac{1}{2} \times 6 \times 4) = 12

sq. units.

Page No 87:

Question 14:

Solve the following system of linear equations graphically:
$x - y + 1 = 0, 3 x + 2 y - 12 = 0 .$
Calculate the area bounded by these lines and the x-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) .........(i)
Putting x = −1, we get y = 0.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.

x	−1	1	2
y	0	2	3

Now, plot the points A(-1, 0), B( 1, 2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of x − y + 1 = 0.

Graph of 3x + 2y − 12 = 0
3x + 2y − 12 = 0
⇒ 2y = (−3x + 12)
∴

y = \frac{- 3 x + 12}{2}

............(ii)
Putting x = 0, we get y = 6.
Putting x = 2, we get y = 3.
Putting x = 4, we get y = 0.
Thus, we have the following table for the equation 3x + 2y − 12 = 0.

x	0	2	4
y	6	3	0

Now, plot the points P(0, 6) and Q(4 , 0). The point C(2, 3) has already been plotted. Join PC and CQ to get the graph line PQ. Extend it on both ways.
Then, PQ is the graph of the equation 3x + 2y − 12 = 0.

The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔACQ formed by these two lines and the x-axis are Q(4, 0), C(2, 3) and A(−1, 0).
Now, consider ΔACQ.
Here, height = 3 units and base (AQ) = 5 units
Area of ΔACQ =

\frac{1}{2} \times base \times height

sq. units
=

(\frac{1}{2} \times 5 \times 3) = 7.5

sq. units.

Page No 87:

Question 15:

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the x-axis:
$x - 2 y + 2 = 0 2 x + y - 6 = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{x + 2}{2} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 2, y = \frac{- 2 + 2}{2} = 0 For x = 2, y = \frac{2 + 2}{2} = 2 For x = 4, y = \frac{4 + 2}{2} = 3$
Thus, the table for the first equation (x − 2y + 2 = 0) is

x	−2	2	4
y	0	2	3

Now, plot the points A(−2,0), B(2,2) and C(4,3) on a graph paper and join
A, B and C to get the graph of x − 2y + 2 = 0.
From the second equation, write y in terms of x

y = 6 - 2 x . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 1, y = 6 - 2 = 4 For x = 3, y = 0 For x = 4, y = 6 - 8 = - 2

So, the table for the second equation (2x + y − 6 = 0 ) is

x	1	3	4
y	4	0	−2

Now, plot the points D(1,4), E(3,0) and F(4,−2) on the same graph paper and join
D, E and F to get the graph of 2x + y − 6 = 0.

From the graph it is clear that, the given lines intersect at (2,2).
So, the solution of the given system of equations is (2,2).
From the graph, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0).
Now, draw a perpendicular from the intersection point B on the x-axis. So,

Area (∆ B A E) = \frac{1}{2} \times A E \times B M = \frac{1}{2} \times 5 \times 2 = 5 sq . units

Hence, the vertices of the triangle formed by the given lines and the x-axis are
(−2,0), (2,2) and (3,0) and the area of the triangle is 5 sq. units.

Page No 87:

Question 16:

Solve the following system of equations graphically and find the vertices and area of the
triangle formed by these lines and the y-axis:
$2 x - 3 y + 6 = 0 2 x + 3 y - 18 = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{2 x + 6}{3} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 3, y = \frac{- 6 + 6}{3} = 0 For x = 0, y = \frac{0 + 6}{3} = 2 For x = 3, y = \frac{6 + 6}{3} = 4$
Thus, the table for the first equation (2x − 3y + 6 = 0) is

x	−3	0	3
y	0	2	4

Now, plot the points A(−3,0), B(0,2) and C(3,4) on a graph paper and join
A, B and C to get the graph of 2x − 3y + 6 = 0.
From the second equation, write y in terms of x

y = \frac{18 - 2 x}{3} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \frac{18 - 0}{3} = 6 For x = 3, y = \frac{18 - 6}{3} = 4 For x = 9, y = \frac{18 - 18}{3} = 0

So, the table for the second equation (2x + 3y − 18 = 0 ) is

x	0	3	9
y	6	4	0

Now, plot the points D(0,6), E(3,4) and F(9,0) on the same graph paper and join
D, E and F to get the graph of 2x + 3y − 18 = 0.

From the graph it is clear that, the given lines intersect at (3,4).
So, the solution of the given system of equations is (3,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4).
Now, draw a perpendicular from the intersection point E (or C) on the y-axis. So,

Area (∆ E D B) = \frac{1}{2} \times B D \times E M = \frac{1}{2} \times 4 \times 3 = 6 sq . units

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,2), (0,6) and (3,4) and its area is 6 sq. units.

Page No 87:

Question 17:

Solve the following system of equations graphically and find the vertices and
area of the triangle formed by these lines and the y-axis.
$4 x - y - 4 = 0 3 x + 2 y - 14 = 0$

Answer:

From the first equation, write y in terms of x
$y = 4 x - 4 . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = 0, y = 0 - 4 = - 4 For x = 1, y = 4 - 4 = 0 For x = 2, y = 8 - 4 = 4$
Thus, the table for the first equation (4x − y − 4 = 0) is

x	0	1	2
y	−4	0	4

Now, plot the points A(0,−4), B(1,0) and C(2,4) on a graph paper and join
A, B and C to get the graph of 4x − y − 4 = 0.
From the second equation, write y in terms of x

y = \frac{14 - 3 x}{2} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \frac{14 - 0}{2} = 7 For x = 4, y = \frac{14 - 12}{2} = 1 For x = \frac{14}{3}, y = \frac{14 - 14}{2} = 0

So, the table for the second equation (3x + 2y − 14 = 0 ) is

x	0	4	$\frac{14}{3}$
y	7	1	0

Now, plot the points D(0,7), E(4,1) and

F (\frac{14}{3}, 0)

on the same graph paper and join
D, E and F to get the graph of 3x + 2y − 14 = 0.

From the graph it is clear that, the given lines intersect at (,4).
So, the solution of the given system of equations is (2,4).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4).
Now, draw a perpendicular from the intersection point C on the y-axis. So,

Area (∆ D A C) = \frac{1}{2} \times D A \times C M = \frac{1}{2} \times 11 \times 2 = 11 sq . units

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,7), (0, −4) and (2,4) and its area is 11 sq. units.

Page No 87:

Question 18:

Solve the following system of equations graphically and find the vertices
and area of the triangle formed by these lines and the y-axis:
$x - y - 5 = 0 3 x + 5 y - 15 = 0$

Answer:

From the first equation, write y in terms of x
$y = x - 5 . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = 0, y = 0 - 5 = - 5 For x = 2, y = 2 - 5 = - 3 For x = 5, y = 5 - 5 = 0$
Thus, the table for the first equation (x − y − 5 = 0) is

x	0	2	5
y	−5	−3	0

Now, plot the points A(0,−5), B(2,−3) and C(5,0) on a graph paper and join
A, B and C to get the graph of x − y − 5 = 0.
From the second equation, write y in terms of x

y = \frac{15 - 3 x}{5} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 5, y = \frac{15 + 15}{5} = 6 For x = 0, y = \frac{15 - 0}{5} = 3 For x = 5, y = \frac{15 - 15}{5} = 0

So, the table for the second equation (3x + 5y − 15 = 0 ) is

x	−5	0	5
y	6	3	0

Now, plot the points D(−5,6), E(0,3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x + 5y − 15 = 0.

From the graph it is clear that, the given lines intersect at (5,0).
So, the solution of the given system of equations is (5,0).
From the graph, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0).
Now,

Area (∆ C E A) = \frac{1}{2} \times E A \times O C = \frac{1}{2} \times 8 \times 5 = 20 sq . units

Hence, the vertices of the triangle formed by the given lines and the y-axis are
(0,3), (0, −5) and (5,0) and its area is 20 sq. units.

Page No 87:

Question 19:

Solve the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis.
$2 x - 5 y + 4 = 0 2 x + y - 8 = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{2 x + 4}{5} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 2, y = \frac{- 4 + 4}{5} = 0 For x = 0, y = \frac{0 + 4}{5} = \frac{4}{5} For x = 3, y = \frac{6 + 4}{5} = 2$
Thus, the table for the first equation (2x − 5y + 4 = 0) is

x	−2	0	3
y	0	$\frac{4}{5}$	2

Now, plot the points A(−2,0),

B (0, \frac{4}{5})

and C(3,2) on a graph paper and join
A, B and C to get the graph of 2x − 5y + 4 = 0.
From the second equation, write y in terms of x

y = 8 - 2 x . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = 8 - 0 = 8 For x = 2, y = 8 - 4 = 4 For x = 4, y = 8 - 8 = 0

So, the table for the second equation (2x + y − 8 = 0 ) is

x	0	2	4
y	8	4	0

Now, plot the points D(0,8), E(2,4) and F(4,0) on the same graph paper and join
D, E and F to get the graph of 2x + y − 8 = 0.

From the graph it is clear that, the given lines intersect at (3,2).
So, the solution of the given system of equations is (3,2).
The vertices of the triangle formed by the system of equations and y-axis are (0,8),

(0, \frac{4}{5})

and (3,2).
Draw a perpendicular from point C to the y-axis. So,

Area (∆ D B C) = \frac{1}{2} \times D B \times C M = \frac{1}{2} \times (8 - \frac{4}{5}) \times 3 = \frac{54}{5} sq . units

Hence, the veritices of the triangle are (0,8),

(0, \frac{4}{5})

and (3,2) and its area is

\frac{54}{5}

sq. units.

Page No 87:

Question 20:

Solve graphically the system of equations:
$5 x - y = 7, x - y + 1 = 0 .$
Calculate the area bounded by these lines and the y-axis.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 5x − y = 7
5x − y = 7
⇒ y = (5x − 7) .........(i)
Putting x = 0, we get y = −7.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation 5x − y = 7.

x	0	1	2
y	−7	−2	3

Now, plot the points A(0, −7) , B( 1, −2) and C(2, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 5x − y = 7.

Graph of x − y + 1 = 0
x − y + 1 = 0
⇒ y = (x + 1) ...........(ii)
Putting x = 0, we get y = 1.
Putting x = 1, we get y = 2.
Putting x = 2, we get y = 3.
Thus, we have the following table for the equation x − y + 1 = 0.

x	0	1	2
y	1	2	3

Now, plot the points P(0, 1) and Q(1, 2). The point C(2, 3) has already been plotted. Join PQ and QC to get the graph line PC. Extend it on both ways.
Then, PC is the graph of the equation x − y + 1 = 0.

The two graph lines intersect at C(2, 3).
∴ The solution of the given system of equations is x = 2 and y = 3.
Clearly, the vertices of ΔAPC formed by these two lines and the y-axis are P(0, 1), C(2, 3) and A(0, −7).
Now, consider ΔAPC.
Here, height = 2 units and base (AP) = 8 units
∴ Area of ΔAPC =

\frac{1}{2} \times base \times height

sq. units
=

(\frac{1}{2} \times 8 \times 2) = 8

sq. units.

Page No 87:

Question 21:

Solve the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the y-axis.
$2 x - 3 y = 12 x + 3 y = 6$

Answer:

From the first equation, write y in terms of x
$y = \frac{2 x - 12}{3} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = 0, y = \frac{0 - 12}{3} = - 4 For x = 3, y = \frac{6 - 12}{3} = - 2 For x = 6, y = \frac{12 - 12}{3} = 0$
Thus, the table for the first equation (2x − 3y = 12) is

x	0	3	6
y	−4	−2	0

Now, plot the points A(0,−4), B(3,−2) and C(6,0) on a graph paper and join
A, B and C to get the graph of 2x − 3y = 12.
From the second equation, write y in terms of x

y = \frac{6 - x}{3} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \frac{6 - 0}{3} = 2 For x = 3, y = \frac{6 - 3}{3} = 1 For x = 6, y = \frac{6 - 6}{3} = 0

So, the table for the second equation (x + 3y = 6 ) is

x	0	3	6
y	2	1	0

Now, plot the points D(0,2), E(3,1) and F(6,0) on the same graph paper and join
D, E and F to get the graph of x + 3y = 6.

From the graph it is clear that, the given lines intersect at (6,0).
So, the solution of the given system of equations is (6,0).
The vertices of the triangle formed by the system of equations and y-axis are (0,2), (6,0) and (0,−4).
Now,

Area (∆ D A C) = \frac{1}{2} \times D A \times O C = \frac{1}{2} \times 6 \times 6 = 18 sq . units

Hence, the veritices of the triangle are (0,2), (6,0) and (0,−4) and its area is 18 sq. units.

Page No 88:

Question 22:

Show graphically that the following given systems of equations has infinitely many solutions:
$2 x + 3 y = 6 4 x + 6 y = 12$

Answer:

From the first equation, write y in terms of x
$y = \frac{6 - 2 x}{3} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 3, y = \frac{6 + 6}{3} = 4 For x = 3, y = \frac{6 - 6}{3} = 0 For x = 6, y = \frac{6 - 12}{3} = - 2$
Thus, the table for the first equation (2x + 3y = 6) is

x	−3	3	6
y	4	0	−2

Now, plot the points A(−3,4), B(3,0) and C(6,−2) on a graph paper and join
A, B and C to get the graph of 2x + 3y = 6.
From the second equation, write y in terms of x

y = \frac{12 - 4 x}{6} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 6, y = \frac{12 + 24}{6} = 6 For x = 0, y = \frac{12 - 0}{6} = 2 For x = 9, y = \frac{12 - 36}{6} = - 4

So, the table for the second equation (4x + 6y = 12 ) is

x	−6	0	9
y	6	2	−4

Now, plot the points D(−6,6), E(0,2) and F(9,−4) on the same graph paper and join
D, E and F to get the graph of 4x + 6y = 12.

From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.

Page No 88:

Question 23:

Show graphically that the system of equations $3 x - y = 5, 6 x - 2 y = 10$ has infinitely many solutions.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 3x − y = 5
3x − y = 5
⇒ y = (3x − 5) .........(i)
Putting x = 1, we get y = −2.
Putting x = 0, we get y = −5.
Putting x = 2, we get y = 1.
Thus, we have the following table for the equation 3x − y = 5.

x	1	0	2
y	−2	−5	1

Now, plot the points A(1, −2), B(0, −5) and C(2, 1) on the graph paper.
Join AB and AC to get the graph line BC. Extend it on both ways.
Thus, BC is the graph of 3x − y = 5.

Graph of 6x − 2y = 10
6x − 2y = 10
⇒ 2y = (6x − 10)
∴

y = \frac{6 x - 10}{2}

...........(ii)
Putting x = 0, we get y = −5.
Putting x = 1, we get y = −2.
Putting x = 2, we get y = 1.

Thus, we have the following table for the equation 6x − 2y = 10.

x	0	1	2
y	−5	−2	1

These are the same points as obtained for the graph line of equation (i).

It is clear from the graph that these two lines coincide.
Hence, the given system of equations has infinitely many solutions.

Page No 88:

Question 24:

Show graphically that the system of equations $2 x + y = 6, 6 x + 3 y = 18$ has infinitely many solutions.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + y = 6
2x + y = 6 ⇒ y = (6 − 2x) ...(i)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 2x + y = 6.

x	3	1	2
y	0	4	2

Now, plot the points A(3, 0), B(1, 4) and C(2, 2) on the graph paper.
Join AC and CB to get the graph line AB. Extend it on both ways.
Thus, AB is the graph of 2x + y = 6.

Graph of 6x + 3y = 18
6x + 3y = 18 ⇒ 3y = (18 − 6x)
∴

y = \frac{18 - 6 x}{3}

...........(ii)
Putting x = 3, we get y = 0.
Putting x = 1, we get y = 4.
Putting x = 2, we get y = 2.
Thus, we have the following table for the equation 6x + 3y = 18.

x	3	1	2
y	0	4	2

These are the same points as obtained for the graph line of equation (i).

It is clear from the graph that these two lines coincide.
Hence, the given system of equations has an infinite number of solutions.

Page No 88:

Question 25:

Show graphically that the following given systems of equations has infinitely many solutions:
$x - 2 y = 5 3 x - 6 y = 15$

Answer:

From the first equation, write y in terms of x
$y = \frac{x - 5}{2} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 5, y = \frac{- 5 - 5}{2} = - 5 For x = 1, y = \frac{1 - 5}{2} = - 2 For x = 3, y = \frac{3 - 5}{2} = - 1$
Thus, the table for the first equation (x − 2y = 5) is

x	−5	1	3
y	−5	−2	−1

Now, plot the points A(−5,−5), B(1,−2) and C(3,−1) on a graph paper and join
A, B and C to get the graph of x − 2y = 5.
From the second equation, write y in terms of x

y = \frac{3 x - 15}{6} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 3, y = \frac{- 9 - 15}{6} = - 4 For x = - 1, y = \frac{- 3 - 15}{6} = - 3 For x = 5, y = \frac{15 - 15}{6} = 0

So, the table for the second equation (3x − 6y = 15 ) is

x	−3	−1	5
y	−4	−3	0

Now, plot the points D(−3,−4), E(−1,−3) and F(5,0) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 15.

From the graph it is clear that, the given lines coincide with each other.
Hence, the solution of the given system of equations has infinitely many solutions.

Page No 88:

Question 26:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:
$x - 2 y = 6 3 x - 6 y = 0$

Answer:

From the first equation, write y in terms of x
$y = \frac{x - 6}{2} . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = - 2, y = \frac{- 2 - 6}{2} = - 4 For x = 0, y = \frac{0 - 6}{2} = - 3 For x = 2, y = \frac{2 - 6}{2} = - 2$
Thus, the table for the first equation (x − 2y = 6) is

x	−2	0	2
y	−4	−3	−2

Now, plot the points A(−2,−4), B(0,−3) and C(2,−2) on a graph paper and join
A, B and C to get the graph of x − 2y = 6.
From the second equation, write y in terms of x

y = \frac{1}{2} x . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = - 4, y = \frac{- 4}{2} = - 2 For x = 0, y = \frac{0}{2} = 0 For x = 4, y = \frac{4}{2} = 2

So, the table for the second equation (3x − 6y = 0 ) is

x	−4	0	4
y	−2	0	2

Now, plot the points D(−4,−2), O(0,0) and E(4,2) on the same graph paper and join
D, E and F to get the graph of 3x − 6y = 0.

From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.

Page No 88:

Question 27:

Show graphically that the system of equations $2 x + 3 y = 4, 4 x + 6 y = 12$ is inconsistent.

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and y-axis, respectively.
Graph of 2x + 3y = 4
2x + 3y = 4
⇒ 3y = (−2x + 4)
∴ $y = \frac{- 2 x + 4}{3}$ .........(i)
Putting x = 2, we get y = 0.
Putting x = −1, we get y = 2.
Putting x = −4, we get y = 4.
Thus, we have the following table for the equation 2x + 3y = 4.

x	2	−1	−4
y	0	2	4

Now, plot the points A(2, 0), B(−1, 2) and C(−4, 4) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both ways.
Thus, AC is the graph of 2x + 3y = 4.

Graph of 4x + 6y = 12
4x + 6y = 12
⇒ 6y = (−4x + 12)
∴

y = \frac{- 4 x + 12}{6}

...........(ii)
Putting x = 3, we get y = 0.
Putting x = 0, we get y = 2.
Putting x = 6, we get y = −2.
Thus, we have the following table for the equation 4x + 6y = 12.

x	3	0	6
y	0	2	−2

Now, on the same graph, plot the points P(3, 0), Q(0, 2) and R (6, −2).
Join PQ and PR to get the graph line QR. Extend it on both ways.
Then, QR is the graph of the equation 4x + 6y = 12.

It is clear from the graph that these two lines are parallel and do not intersect when produced.
Hence, the given system of equations is inconsistent.

Page No 88:

Question 28:

Show graphically that the following given systems of equations is inconsistent i.e. has no solution:
$2 x + y = 6 6 x + 3 y = 20$

Answer:

From the first equation, write y in terms of x
$y = 6 - 2 x . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = 0, y = 6 - 0 = 6 For x = 2, y = 6 - 4 = 2 For x = 4, y = 6 - 8 = - 2$
Thus, the table for the first equation (2x + y = 6) is

x	0	2	4
y	6	2	−2

Now, plot the points A(0,6), B(2,2) and C(4,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 6.
From the second equation, write y in terms of x

y = \frac{20 - 6 x}{3} . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = \frac{20 - 0}{3} = \frac{20}{3} For x = \frac{10}{3}, y = \frac{20 - 20}{3} = 0 For x = 5, y = \frac{20 - 30}{3} = - \frac{10}{3}

So, the table for the second equation (6x + 3y = 20 ) is

x	0	$\frac{10}{3}$	5
y	$\frac{20}{3}$	0	$- \frac{10}{3}$

Now, plot the points

D (0, \frac{20}{3}), E (\frac{10}{3}, 0) and F (5, - \frac{10}{3})

on the same graph paper and join
D, E and F to get the graph of 6x + 3y = 20.

From the graph it is clear that, the given lines do not intersect at all when produced.
Hence, the system of equations has no solution and therefore is inconsistent.

Page No 88:

Question 29:

Draw the graphs of the following equations on the same graph paper:
$2 x + y = 2 2 x + y = 6$
Find the coordinates of the vertices of the trapezium formed by these lines. Also, find the area of the trapezium so formed.

Answer:

From the first equation, write y in terms of x
$y = 2 - 2 x . . . . . (i)$
Substitute different values of x in (i) to get different values of y
$For x = 0, y = 2 - 0 = 2 For x = 1, y = 2 - 2 = 0 For x = 2, y = 2 - 4 = - 2$
Thus, the table for the first equation (2x + y = 2) is

x	0	1	2
y	2	0	−2

Now, plot the points A(0,2), B(1,0) and C(2,−2) on a graph paper and join
A, B and C to get the graph of 2x + y = 2.
From the second equation, write y in terms of x

y = 6 - 2 x . . . . . (i i)

Now, substitute different values of x in (ii) to get different values of y

For x = 0, y = 6 - 0 = 6 For x = 1, y = 6 - 2 = 4 For x = 3, y = 6 - 6 = 0

So, the table for the second equation (2x + y = 6 ) is

x	0	1	3
y	6	4	0

Now, plot the points D(0,6), E(1,4) and F(3,0)on the same graph paper and join
D, E and F to get the graph of 2x + y = 6.

From the graph it is clear that, the given lines do not intersect at all when produced. So, these lines are
parallel to each other and therefore, the quadrilateral DABF is a trapezium. The vertices of the
required trapezium are D(0,6), A(0,2), B(1,0) and F(3,0).
Now,

Area (T r a p e z i u m D A B F) = Area (∆ D O F) - Area (∆ A O B) = \frac{1}{2} \times 3 \times 6 - \frac{1}{2} \times 1 \times 2 = 9 - 1 = 8 sq . units

Hence, the area of the rquired trapezium is 8 sq. units.

Page No 103:

Question 1:

$x + y = 3, 4 x - 3 y = 26 .$

Answer:

The given system of equation is:
x + y = 3 .........(i)
4x − 3y = 26 ........(ii)

On multiplying (i) by 3, we get:
3x + 3y = 9 .......(iii)

On adding (ii) and (iii), we get:
7x = 35
⇒ x = 5

On substituting the value of x = 5 in (i), we get:
5 + y = 3
⇒ y = (3 − 5) = −2

Hence, the solution is x = 5 and y = −2

Page No 103:

Question 2:

Solve for x and y:
$x - y = 3 \frac{x}{3} + \frac{y}{2} = 6$

Answer:

The given system of equations is
$x - y = 3 . . . . . (i) \frac{x}{3} + \frac{y}{2} = 6 . . . . . (i i)$
From (i), write y in terms of x to get
$y = x - 3$
Substituting y = x − 3 in (ii), we get
$\frac{x}{3} + \frac{x - 3}{2} = 6 \Rightarrow 2 x + 3 (x - 3) = 36 \Rightarrow 2 x + 3 x - 9 = 36 \Rightarrow x = \frac{45}{5} = 9$
Now, substituting x = 9 in (i), we have
$9 - y = 3 \Rightarrow y = 9 - 3 = 6$
Hence, x = 9 and y = 6.

Page No 104:

Question 3:

$2 x + 3 y = 0, 3 x + 4 y = 5 .$

Answer:

The given system of equation is:
2x + 3y = 0 .........(i)
3x + 4y = 5 ........(ii)

On multiplying (i) by 4 and (ii) by 3, we get:
8x + 12y = 0 .......(iii)
9x + 12y = 15 ......(iv)

On subtracting (iii) from (iv) we get:
x = 15

On substituting the value of x = 15 in (i), we get:
30 + 3y = 0
⇒ 3y = −30
⇒ y = −10

Hence, the solution is x = 15 and y = −10.

Page No 104:

Question 4:

$2 x - 3 y = 13, 7 x - 2 y = 20 .$

Answer:

The given system of equation is:
2x − 3y = 13 .........(i)
7x − 2y = 20 ........(ii)

On multiplying (i) by 2 and (ii) by 3, we get:
4x − 6y = 26 .......(iii)
21x − 6y = 60 ......(iv)

On subtracting (iii) from (iv) we get:
17x = (60 − 26) = 34
⇒ x = 2

On substituting the value of x = 2 in (i), we get:
4 − 3y = 13
⇒ 3y = (4 − 13) = −9
⇒ y = −3

Hence, the solution is x = 2 and y = −3

Page No 104:

Question 5:

$3 x - 5 y - 19 = 0, - 7 x + 3 y + 1 = 0 .$

Answer:

The given system of equation is:
3x − 5y − 19 = 0 .........(i)
−7x + 3y + 1 = 0 ........(ii)

On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 57 or .........(iii)
−35x + 15y = −5 ........(iv)

On adding (iii) from (iv) we get:
−26x = (57 − 5)) = 52
⇒ x = −2

On substituting the value of x = −2 in (i), we get:
−6 − 5y − 19 = 0
⇒ 5y = (− 6 − 19) = −25
⇒ y = −5

Hence, the solution is x = −2 and y = −5.

Page No 104:

Question 6:

Solve for x and y:
$2 x - y + 3 = 0 3 x - 7 y + 10 = 0$

Answer:

The given system of equations is
$2 x - y + 3 = 0 . . . . . (i) 3 x - 7 y + 10 = 0 . . . . . (i i)$
From (i), write y in terms of x to get
y = 2x + 3
Substituting y = 2x + 3 in (ii), we get
$3 x - 7 (2 x + 3) + 10 = 0 \Rightarrow 3 x - 14 x - 21 + 10 = 0 \Rightarrow - 7 x = 21 - 10 = 11 \Rightarrow x = - \frac{11}{7}$
Now, substituting $x = - \frac{11}{7}$ in (i), we have
$- \frac{22}{7} - y + 3 = 0 \Rightarrow y = 3 - \frac{22}{7} = - \frac{1}{7}$
Hence, $x = - \frac{11}{7}$ and $y = - \frac{1}{7}$ .

Page No 104:

Question 7:

Solve for x and y:
$\frac{x}{2} - \frac{y}{9} = 6 \frac{x}{7} + \frac{y}{3} = 5$

−117 y=−1

Answer:

The given system of equations can be written as
$9 x - 2 y = 108 . . . . . (i) 3 x + 7 y = 105 . . . . . (i i)$
Multiplying (i) by 7 and (ii) by 2, we get
$63 x + 6 x = 108 \times 7 + 105 \times 2 \Rightarrow 69 x = 966 \Rightarrow x = \frac{966}{69} = 14$
Now, substituting x = 14 in (1), we have
$9 \times 14 - 2 y = 108 \Rightarrow 2 y = 126 - 108 \Rightarrow y = \frac{18}{2} = 9$
Hence, x = 14 and y = 9.

Page No 104:

Question 8:

$\frac{x}{3} + \frac{y}{4} = 11, \frac{5 x}{6} - \frac{y}{3} = - 7 .$

Answer:

The given equations are:
$\frac{x}{3} + \frac{y}{4} = 11$
⇒ 4x + 3y = 132 ........(i)

and $\frac{5 x}{6} - \frac{y}{3} = - 7$
⇒ 5x − 2y = −42..........(ii)

On multiplying (i) by 2 and (ii) by 3, we get:
8x + 6y = 264 ........(iii)
15x − 6y = −126 ...........(iv)

On adding (iii) and (iv), we get:
23x = 138
⇒ x = 6

On substituting x = 6 in (i), we get:
24 + 3y = 132
⇒ 3y = (132 − 24) = 108
⇒ y = 36

Hence, the solution is x = 6 and y = 36.

Page No 104:

Question 9:

$4 x - 3 y = 8, 6 x - y = \frac{29}{3} .$

Answer:

The given system of equation is:
4x − 3y = 8 .........(i)
$6 x - y = \frac{29}{3}$ ........(ii)
On multiplying (ii) by 3, we get:
18x − 3y = 29.........(iii)

On subtracting (iii) from (i) we get:
−14x = −21
⇒ x = $\frac{21}{14} = \frac{3}{2}$
On substituting the value of x = $\frac{3}{2}$ in (i), we get:
$4 \times \frac{3}{2} - 3 y = 8 \Rightarrow 6 - 3 y = 8 \Rightarrow 3 y = 6 - 8 = - 2 \Rightarrow y = \frac{- 2}{3}$
Hence, the solution is x = $\frac{3}{2}$ and y = $\frac{- 2}{3}$ .

Page No 104:

Question 10:

$2 x - \frac{3 y}{4} = 3, 5 x = 2 y + 7 .$

Answer:

The given equations are:
$2 x - \frac{3 y}{4} = 3$ ........(i)
5x = 2y + 7 ............(ii)

On multiplying (i) by 2 and (ii) by $\frac{3}{4}$ , we get:
$4 x - \frac{3}{2} y = 6$ .......(iii)
$\frac{15}{4} x = \frac{3}{2} y + \frac{21}{4}$ .......(iv)

On subtracting (iii) from (iv), we get:
$- \frac{1}{4} x = - \frac{3}{4} \Rightarrow x = 3$

On substituting x = 3 in (i), we get:
$2 \times 3 - \frac{3 y}{4} = 3 \Rightarrow \frac{3 y}{4} = (6 - 3) = 3 \Rightarrow y = \frac{3 \times 4}{3} = 4$

Hence, the solution is x = 3 and y = 4.

Page No 104:

Question 11:

$2 x + 5 y = \frac{8}{3}, 3 x - 2 y = \frac{5}{6} .$

Answer:

The given equations are:
$2 x + 5 y = \frac{8}{3}$ ........(i)
$3 x - 2 y = \frac{5}{6}$ ..........(ii)

On multiplying (i) by 2 and (ii) by 5, we get:
$4 x + 10 y = \frac{16}{3}$ ........(iii)
$15 x - 10 y = \frac{25}{6}$ ...........(iv)

On adding (iii) and (iv), we get:
$19 x = \frac{57}{6} \Rightarrow x = \frac{57}{6 \times 19} = \frac{3}{6} = \frac{1}{2}$

On substituting x = $\frac{1}{2}$ in (i), we get:
$2 \times \frac{1}{2} + 5 y = \frac{8}{3}$
$\Rightarrow 5 y = (\frac{8}{3} - 1) = \frac{5}{3} \Rightarrow y = \frac{5}{3 \times 5} = \frac{1}{3}$

Hence, the solution is x = $\frac{1}{2}$ and y = $\frac{1}{3}$ .

Page No 104:

Question 12:

$\frac{7 - 4 x}{3} = y, 2 x + 3 y + 1 = 0 .$

Answer:

The given equations are:
$\frac{7 - 4 x}{3} = y$
⇒ 4x + 3y = 7 .......(i)

and 2x + 3y + 1 = 0
⇒ 2x + 3y = −1 .............(ii)

On subtracting (ii) from (i), we get:
2x = 8
⇒ x = 4

On substituting x = 4 in (i), we get:
16 + 3y = 7
⇒ 3y = (7 − 16) = −9
⇒ y = −3

Hence, the solution is x = 4 and y = −3.

Page No 104:

Question 13:

Solve for x and y:
$0.4 x + 0.3 y = 1.7 0.7 x - 0.2 y = 0.8$

Answer:

The given system of equations is
$0.4 x + 0.3 y = 1.7 . . . . . (i) 0.7 x - 0.2 y = 0.8 . . . . . (i i)$
Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get
$0.8 x + 2.1 x = 3.4 + 2.4 \Rightarrow 2.9 x = 5.8 \Rightarrow x = \frac{5.8}{2.9} = 2$
Now, substituting x = 2 in (i), we have
$0.4 \times 2 + 0.3 y = 1.7 \Rightarrow 0.3 y = 1.7 - 0.8 \Rightarrow y = \frac{0.9}{0.3} = 3$
Hence, x = 2 and y = 3.

Page No 104:

Question 14:

Solve for x and y:
$0.3 x + 0.5 y = 0.5 0.5 x + 0.7 y = 0.74$

Answer:

The given system of equations is
$0.3 x + 0.5 y = 0.5 . . . . . (i) 0.5 x + 0.7 y = 0.74 . . . . . (i i)$
Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get
$2.5 y - 2.1 y = 2.5 - 2.22 \Rightarrow 0.4 y = 0.28 \Rightarrow y = \frac{0.28}{0.4} = 0.7$
Now, substituting y = 0.7 in (i), we have
$0.3 x + 0.5 \times 0.7 = 0.5 \Rightarrow 0.3 x = 0.50 - 0.35 = 0.15 \Rightarrow x = \frac{0.15}{0.3} = 0.5$
Hence, x = 0.5 and y = 0.7.

Page No 104:

Question 15:

$7 (y + 3) - 2 (x + 2) = 14, 4 (y - 2) + 3 (x - 3) = 2 .$

Answer:

The given equations are:
7(y + 3) − 2(x + 2) = 14
⇒ 7y + 21 − 2x − 4 = 14
⇒ −2x + 7y = −3 ..........(i)

and 4(y − 2) + 3(x − 3) = 2
⇒ 4y − 8 + 3x − 9 = 2
⇒ 3x + 4y = 19 .........(ii)

On multiplying (i) by 4 and (ii) by 7, we get:
−8x + 28y = −12 ........(iii)
21x + 28y = 133 ...........(iv)

On subtracting (iii) from (iv), we get:
29x = 145
⇒ x = 5

On substituting x = 5 in (i), we get:
−10 + 7y = −3
⇒ 7y = (−3 + 10) = 7
⇒ y = 1

Hence, the solution is x = 5 and y = 1.

Page No 104:

Question 16:

$6 x + 5 y = 7 x + 3 y + 1 = 2 (x + 6 y - 1)$

Answer:

The given equations are:
6x + 5y = 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 6x + 5y = 2(x + 6y − 1)
⇒ 6x + 5y = 2x + 12y − 2
⇒ 6x − 2x + 5y − 12y = − 2
⇒ 4x − 7y = −2 ..........(i)

and 7x + 3y + 1 = 2(x + 6y − 1)
⇒ 7x + 3y + 1 = 2x + 12y − 2
⇒ 7x − 2x + 3y − 12y = −2 − 1
⇒ 5x − 9y = −3 .........(ii)

On multiplying (i) by 9 and (ii) by 7, we get:
36x − 63y = −18 ..........(iii)
35x − 63y = −21.........(iv)

On subtracting (iv) from (iii), we get:
x = (−18 + 21) = 3

On substituting x = 3 in (i), we get:
12 − 7y = −2
⇒ 7y = (2 + 12) = 14
⇒ y = 2

Hence, the solution is x = 3 and y = 2.

Page No 104:

Question 17:

$\frac{x + y - 8}{2} = \frac{x + 2 y - 14}{3} = \frac{3 x + y - 12}{11}$

Answer:

The given equations are:
$\frac{x + y - 8}{2} = \frac{x + 2 y - 14}{3} = \frac{3 x + y - 12}{11}$
i.e., $\frac{x + y - 8}{2} = \frac{3 x + y - 12}{11}$
By cross multiplication, we get:
11x + 11y − 88 = 6x + 2y − 24
⇒ 11x − 6x + 11y − 2y = −24 + 88
⇒ 5x + 9y = 64 .........(i)

and $\frac{x + 2 y - 14}{3} = \frac{3 x + y - 12}{11}$
⇒ 11x + 22y − 154 = 9x + 3y − 36
⇒ 11x − 9x + 22y − 3y = −36 + 154
⇒ 2x + 19y = 118 .........(ii)

On multiplying (i) by 19 and (ii) by 9, we get:
95x + 171y = 1216..........(iii)
18x + 171y = 1062.........(iv)

On subtracting (iv) from (iii), we get:
77x = 154
⇒ x = 2

On substituting x = 2 in (i), we get:
10 + 9y = 64
⇒ 9y = (64 − 10) = 54
⇒ y = 6

Hence, the solution is x = 2 and y = 6.

Page No 104:

Question 18:

$\frac{5}{x} + 6 y = 13, \frac{3}{x} + 4 y = 7, x \neq 0 .$

Answer:

The given equations are:
$\frac{5}{x} + 6 y = 13$ ............(i)
$\frac{3}{x} + 4 y = 7$ .............(ii)

Putting $\frac{1}{x} = u$ , we get:
5u + 6y = 13 .............(iii)
3u + 4y = 7 ...........(iv)

On multiplying (iii) by 4 and (iv) by 6, we get:
20u + 24y = 52 ...........(v)
18u + 24y = 42 ............(vi)

On subtracting (vi) from (v), we get:
2u = 10 ⇒ u = 5
$\Rightarrow \frac{1}{x} = 5 \Rightarrow x = \frac{1}{5}$
On substituting $x = \frac{1}{5}$ in (i), we get:
$\frac{5}{\frac{1}{5}} + 6 y = 13$
⇒ 25 + 6y = 13
⇒ 6y = (13 − 25) = −12
⇒ y = −2

Hence, the required solution is $x = \frac{1}{5}$ and y = −2.

Page No 104:

Question 19:

$x + \frac{6}{y} = 6, 3 x - \frac{8}{y} = 5, y \neq 0 .$

Answer:

The given equations are:
$x + \frac{6}{y} = 6$ ............(i)
$3 x - \frac{8}{y} = 5$ .............(ii)
Putting $\frac{1}{y} = v$ , we get:
x + 6v = 6 .............(iii)
3x − 8v = 5 ...........(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
4x + 24v = 24 ...........(v)
9x − 24v = 15 ............(vi)

On adding (v) and (vi), we get:
13x = 39 ⇒ x = 3

On substituting x = 3 in (i), we get:
$3 + \frac{6}{y} = 6$
$\Rightarrow \frac{6}{y} = (6 - 3) = 3 \Rightarrow 3 y = 6 \Rightarrow y = 2$
Hence, the required solution is x = 3 and y = 2.

Page No 104:

Question 20:

$2 x - \frac{3}{y} = 9, 3 x + \frac{7}{y} = 2, y \neq 0 .$

Answer:

The given equations are:
$2 x - \frac{3}{y} = 9$ ............(i)
$3 x + \frac{7}{y} = 2$ .............(ii)
Putting $\frac{1}{y} = v$ , we get:
2x − 3v = 9 .............(iii)
3x + 7v = 2 ...........(iv)

On multiplying (iii) by 7 and (iv) by 3, we get:
14x − 21v = 63 .............(v)
9x + 21v = 6............(vi)

On adding (v) and (vi), we get:
23x = 69 ⇒ x = 3

On substituting x = 3 in (i), we get:
$2 \times 3 - \frac{3}{y} = 9$
$\Rightarrow 6 - \frac{3}{y} = 9 \Rightarrow \frac{3}{y} = - 3 \Rightarrow y = - 1$
Hence, the required solution is x = 3 and y = −1.

Page No 104:

Question 21:

$\frac{3}{x} - \frac{1}{y} + 9 = 0, \frac{2}{x} + \frac{3}{y} = 5 (x \neq 0, y \neq 0 .)$

Answer:

The given equations are:
$\frac{3}{x} - \frac{1}{y} + 9 = 0$
⇒ $\frac{3}{x} - \frac{1}{y} = - 9$ ............(i)
$\frac{2}{x} + \frac{3}{y} = 5$ .............(ii)
Putting $\frac{1}{x} = u$ and $\frac{1}{y} = v$ , we get:
3u − v = −9 .............(iii)
2u + 3v = 5 ...........(iv)

On multiplying (iii) by 3, we get:
9u − 3v = −27 .............(v)

On adding (iv) and (v), we get:
11u = −22 ⇒ u = −2
$\Rightarrow \frac{1}{x} = - 2 \Rightarrow x = \frac{- 1}{2}$

On substituting $x = \frac{- 1}{2}$ in (i), we get:
$\frac{3}{- \frac{1}{2}} - \frac{1}{y} = - 9$
$\Rightarrow - 6 - \frac{1}{y} = - 9 \Rightarrow \frac{1}{y} = (- 6 + 9) = 3$
$\Rightarrow y = \frac{1}{3}$
Hence, the required solution is $x = \frac{- 1}{2}$ and $y = \frac{1}{3}$ .

Page No 104:

Question 22:

$\frac{9}{x} - \frac{4}{y} = 8, \frac{13}{x} + \frac{7}{y} = 101 (x \neq 0, y \neq 0) .$

Answer:

The given equations are:
$\frac{9}{x} - \frac{4}{y} = 8$ ............(i)
$\frac{13}{x} + \frac{7}{y} = 101$ .............(ii)

Putting $\frac{1}{x} = u$ and $\frac{1}{y} = v$ , we get:
9u − 4v = 8 .............(iii)
13u + 7v = 101 ...........(iv)

On multiplying (iii) by 7 and (iv) by 4, we get:
63u − 28v = 56 .............(v)
52u + 28v = 404............(vi)

On adding (v) and (vi), we get:
115u = 460 ⇒ u = 4
$\Rightarrow \frac{1}{x} = 4 \Rightarrow x = \frac{1}{4}$
On substituting $x = \frac{1}{4}$ in (i), we get:
$\frac{9}{\frac{1}{4}} - \frac{4}{y} = 8$
$\Rightarrow 36 - \frac{4}{y} = 8 \Rightarrow \frac{4}{y} = (36 - 8) = 28$
$\Rightarrow y = \frac{4}{28} = \frac{1}{7}$
Hence, the required solution is $x = \frac{1}{4}$ and $y = \frac{1}{7}$ .

Page No 104:

Question 23:

$\frac{5}{x} - \frac{3}{y} = 1, \frac{3}{2 x} + \frac{2}{3 y} = 5 (x \neq 0, y \neq 0 .)$

Answer:

The given equations are:
$\frac{5}{x} - \frac{3}{y} = 1$ ............(i)
$\frac{3}{2 x} + \frac{2}{3 y} = 5$ .............(ii)

Putting $\frac{1}{x} = u$ and $\frac{1}{y} = v$ , we get:
5u − 3v = 1 .............(iii)
⇒ $\frac{3}{2} u + \frac{2}{3} v = 5$
$\Rightarrow \frac{9 u + 4 v}{6} = 5$
$\Rightarrow 9 u + 4 v = 30$ ...............(iv)

On multiplying (iii) by 4 and (iv) by 3, we get:
20u − 12v = 4 .............(v)
27u + 12v = 90 ...........(vi)

On adding (iv) and (v), we get:
47u = 94 ⇒ u = 2
$\Rightarrow \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}$

On substituting $x = \frac{1}{2}$ in (i), we get:
$\frac{5}{\frac{1}{2}} - \frac{3}{y} = 1$
$\Rightarrow 10 - \frac{3}{y} = 1 \Rightarrow \frac{3}{y} = (10 - 1) = 9$
$\Rightarrow y = \frac{3}{9} = \frac{1}{3}$
Hence, the required solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$ .

Page No 104:

Question 24:

Solve for x and y:
$\frac{1}{2 x} + \frac{1}{3 y} = 2 \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}$

Answer:

Multiplying equation (i) and (ii) by 6, we get
$\frac{3}{x} + \frac{2}{y} = 12 . . . . . (i) \frac{2}{x} + \frac{3}{y} = 13 . . . . . (i i)$
Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i), we get
$\frac{9}{x} - \frac{4}{x} = 36 - 26 \Rightarrow \frac{5}{x} = 10 \Rightarrow x = \frac{5}{10} = \frac{1}{2}$
Now, substituting $x = \frac{1}{2}$ in (i), we have
$6 + \frac{2}{y} = 12 \Rightarrow \frac{2}{y} = 6 \Rightarrow y = \frac{1}{3}$
Hence, $x = \frac{1}{2}$ and $y = \frac{1}{3}$ .

Page No 104:

Question 25:

$4 x + 6 y = 3 x y, 8 x + 9 y = 5 x y (x \neq 0, y \neq 0) .$

Answer:

The given equations are:
4x + 6y = 3xy .......(i)
8x + 9y = 5xy .........(ii)

From equation (i), we have:

$\frac{4 x + 6 y}{x y} = 3$
$\Rightarrow \frac{4}{y} + \frac{6}{x} = 3$ .............(iii)

For equation (ii), we have:

$\frac{8 x + 9 y}{x y} = 5$
$\Rightarrow \frac{8}{y} + \frac{9}{x} = 5$ .............(iv)
On substituting $\frac{1}{y} = v and \frac{1}{x} = u$ , we get:
4v + 6u = 3 ...........(v)
8v + 9u = 5 ...........(vi)

On multiplying (v) by 9 and (vi) by 6, we get:
36v + 54u = 27 ..............(vii)
48v + 54u = 30 ...............(viii)

On subtracting (vii) from (viii), we get:
12v = 3 $\Rightarrow v = \frac{3}{12} = \frac{1}{4}$
$\Rightarrow \frac{1}{y} = \frac{1}{4} \Rightarrow y = 4$

On substituting y = 4 in (iii), we get:
$\frac{4}{4} + \frac{6}{x} = 3$
$\Rightarrow 1 + \frac{6}{x} = 3 \Rightarrow \frac{6}{x} = (3 - 1) = 2$
$\Rightarrow 2 x = 6 \Rightarrow x = \frac{6}{2} = 3$
Hence, the required solution is x = 3 and y = 4.

Page No 104:

Question 26:

$x + y = 5 x y, 3 x + 2 y = 13 x y .$

Answer:

The given equations are:
x + y = 5xy .......(i)
3x + 2y = 13xy .........(ii)

From equation (i), we have:

$\frac{x + y}{x y} = 5$
$\Rightarrow \frac{1}{y} + \frac{1}{x} = 5$ .............(iii)

From equation (ii), we have:

$\frac{3 x + 2 y}{x y} = 13$
$\Rightarrow \frac{3}{y} + \frac{2}{x} = 13$ .............(iv)
On substituting $\frac{1}{y} = v and \frac{1}{x} = u$ , we get:
v + u = 5 ...........(v)
3v + 2u = 13 ...........(vi)

On multiplying (v) by 2, we get:
2v + 2u = 10 ..............(vii)
On subtracting (vii) from (vi), we get:
v = 3
$\Rightarrow \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3}$
On substituting $y = \frac{1}{3}$ in (iii), we get:
$\frac{1}{\frac{1}{3}} + \frac{1}{x} = 5$
$\Rightarrow 3 + \frac{1}{x} = 5 \Rightarrow \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}$
Hence, the required solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$ or x = 0 and y = 0.

Page No 104:

Question 27:

Solve for x and y:
$\frac{5}{x + y} - \frac{2}{x - y} = - 1 \frac{15}{x + y} + \frac{7}{x - y} = 10$

Answer:

The given equations are
$\frac{5}{x + y} - \frac{2}{x - y} = - 1 . . . . . (i) \frac{15}{x + y} + \frac{7}{x - y} = 10 . . . . . (i i)$
Substituting $\frac{1}{x + y} = u and \frac{1}{x - y} = v$ in (i) and (ii), we get
$5 u - 2 v = - 1 . . . . . (i i i) 15 u + 7 v = 10 . . . . . (i v)$
Multiplying (iii) by 3 and subtracting it from (iv), we get
$7 v + 6 v = 10 + 3 \Rightarrow 13 v = 13 \Rightarrow v = 1 \Rightarrow x - y = 1 (∵ \frac{1}{x - y} = v) . . . . . (v)$
Now, substituting v = 1 in (iii), we get
$5 u - 2 = - 1 \Rightarrow 5 u = 1 \Rightarrow u = \frac{1}{5} x + y = 5 . . . . . (v i)$
Adding (v) and (vi), we get
$2 x = 6 \Rightarrow x = 3$
Substituting x = 3 in (vi), we have
$3 + y = 5 \Rightarrow y = 5 - 3 = 2$
Hence, x = 3 and y = 2.

Page No 105:

Question 28:

$\frac{3}{x + y} + \frac{2}{x - y} = 2, \frac{9}{x + y} - \frac{4}{x - y} = 1, where x + y \neq 0 and x - y \neq 0 .$

Answer:

The given equations are:
$\frac{3}{x + y} + \frac{2}{x - y} = 2$           ...(i)
$\frac{9}{x + y} - \frac{4}{x - y} = 1$            ...(ii)
Putting $\frac{1}{x + y} = u$ and $\frac{1}{x - y} = v$ , we get:
3u + 2v = 2            ...(iii)
9u − 4v = 1              ...(iv)
On multiplying (iii) by 2, we get:
6u + 4v = 4               ...(v)
On adding (iv) and (v), we get:
15u = 5
$\Rightarrow u = \frac{5}{15} = \frac{1}{3}$
$\Rightarrow \frac{1}{x + y} = \frac{1}{3} \Rightarrow x + y = 3$              ...(vi)
On substituting $u = \frac{1}{3}$ in (iii), we get:
1 + 2v = 2
⇒ 2v = 1
$\Rightarrow v = \frac{1}{2}$
$\Rightarrow \frac{1}{x - y} = \frac{1}{2} \Rightarrow x - y = 2$                 ...(vii)
On adding (vi) and (vii), we get:
2x = 5
⇒ $x = \frac{5}{2}$
On substituting $x = \frac{5}{2}$ in (vi), we get:
$\frac{5}{2} + y = 3 \Rightarrow y = (3 - \frac{5}{2}) = \frac{1}{2}$
Hence, the required solution is $x = \frac{5}{2} and y = \frac{1}{2}$ .

Page No 105:

Question 29:

$\frac{5}{x + 1} - \frac{2}{y - 1} = \frac{1}{2}, \frac{10}{x + 1} + \frac{2}{y - 1} = \frac{5}{2}, where x \neq - 1, y \neq 1 .$

Answer:

The given equations are:
$\frac{5}{x + 1} - \frac{2}{y - 1} = \frac{1}{2}$ .............(i)
$\frac{10}{x + 1} + \frac{2}{y - 1} = \frac{5}{2}$ ..............(ii)
Putting $\frac{1}{x + 1} = u$ and $\frac{1}{y - 1} = v$ , we get:
$5 u - 2 v = \frac{1}{2}$ .................(iii)
$10 u + 2 v = \frac{5}{2}$ ................(iv)
On adding (iii) and (iv), we get:
15u = 3
⇒ $u = \frac{3}{15} = \frac{1}{5}$
⇒ $\frac{1}{x + 1} = \frac{1}{5} \Rightarrow x + 1 = 5 \Rightarrow x = 4$
On substituting $u = \frac{1}{5}$ in (iii), we get:
$5 \times \frac{1}{5} - 2 v = \frac{1}{2} \Rightarrow 1 - 2 v = \frac{1}{2}$
$\Rightarrow 2 v = \frac{1}{2} \Rightarrow v = \frac{1}{4}$
$\Rightarrow \frac{1}{y - 1} = \frac{1}{4} \Rightarrow y - 1 = 4 \Rightarrow y = 5$
Hence, the required solution is x = 4 and y = 5.

Page No 105:

Question 30:

$\frac{44}{x + y} + \frac{30}{x - y} = 10, \frac{55}{x + y} + \frac{40}{x - y} = 13 .$

Answer:

The given equations are:
$\frac{44}{x + y} + \frac{30}{x - y} = 10$         ...(i)
$\frac{55}{x + y} + \frac{40}{x - y} = 13$          ...(ii)
Putting $\frac{1}{x + y} = u$ and $\frac{1}{x - y} = v$ , we get:
44u + 30v = 10          ...(iii)
55u + 40v = 13          ...(iv)
On multiplying (iii) by 4 and (iv) by 3, we get:
176u + 120v = 40           ...(v)
165u + 120v = 39           ...(vi)
On subtracting (vi) from (v), we get:
11u = 1
$\Rightarrow u = \frac{1}{11}$
$\Rightarrow \frac{1}{x + y} = \frac{1}{11} \Rightarrow x + y = 11$          ...(vii)
On substituting $u = \frac{1}{11}$ in (iii), we get:
4 + 30v = 10
⇒ 30v = 6
$\Rightarrow v = \frac{6}{30} = \frac{1}{5}$
$\Rightarrow \frac{1}{x - y} = \frac{1}{5} \Rightarrow x - y = 5$      ...(viii)
On adding (vii) and (viii), we get:
2x = 16
⇒ x = 8
On substituting x = 8 in (vii), we get:
8 + y = 11
⇒ y = (11 − 8) = 3
Hence, the required solution is x = 8 and y = 3.

Page No 105:

Question 31:

Solve for x and y:
$\frac{10}{x + y} + \frac{2}{x - y} = 4 \frac{15}{x + y} - \frac{9}{x - y} = - 2$

Answer:

The given equations are
$\frac{10}{x + y} + \frac{2}{x - y} = 4 . . . . . (i) \frac{15}{x + y} - \frac{9}{x - y} = - 2 . . . . . (i i)$
Substituting $\frac{1}{x + y} = u and \frac{1}{x - y} = v$ in (i) and (ii), we get
$10 u + 2 v = 4 . . . . . (i i i) 15 u - 9 v = - 2 . . . . . (i v)$
Multiplying (iii) by 9 and (iv) by 2 and adding, we get
$90 u + 30 u = 36 - 4 \Rightarrow 120 u = 32 \Rightarrow u = \frac{32}{120} = \frac{4}{15} \Rightarrow x + y = \frac{15}{4} (∵ \frac{1}{x + y} = u) . . . . . (v)$
Now, substituting $u = \frac{4}{15}$ in (iii), we get
$10 \times \frac{4}{15} + 2 v = 4 \frac{8}{3} + 2 v = 4 \Rightarrow 2 v = 4 - \frac{8}{3} = \frac{4}{3} \Rightarrow v = \frac{2}{3} \Rightarrow x - y = \frac{3}{2} (∵ \frac{1}{x - y} = v) . . . . . (v i)$
Adding (v) and (vi), we get
$2 x = \frac{15}{4} + \frac{3}{2} \Rightarrow 2 x = \frac{21}{4} \Rightarrow x = \frac{21}{8}$
Substituting $x = \frac{21}{8}$ in (v), we have
$\frac{21}{8} + y = \frac{15}{4} \Rightarrow y = \frac{15}{4} - \frac{21}{8} = \frac{9}{8}$
Hence, $x = \frac{21}{8} and y = \frac{9}{8}$ .

Page No 105:

Question 32:

$71 x + 37 y = 253, 37 x + 71 y = 287 .$

Answer:

The given equations are:
71x + 37y = 253 ..............(i)
37x + 71y = 287 ............(ii)

On adding (i) and (ii), we get:
108x + 108y = 540
⇒ 108(x + y) = 540
⇒ (x + y) = 5................(iii)

On subtracting (ii) from (i), we get:
34x − 34y = −34
⇒ 34(x − y) = −34
⇒ (x − y) = −1...........(iv)

On adding (iii) and (iv), we get:
2x = 5 − 1= 4
⇒ x = 2

On subtracting (iv) from (iii), we get:
2y = 5 + 1 = 6
⇒ y = 3

Hence, the required solution is x = 2 and y = 3.

Page No 105:

Question 33:

$217 x + 131 y = 913, 131 x + 217 y = 827 .$

Answer:

The given equations are:
217x + 131y = 913 .............(i)
131x + 217y = 827 .............(ii)

On adding (i) and (ii), we get:
348x + 348y = 1740
⇒ 348(x + y) = 1740
⇒ x + y = 5 ............(iii)

On subtracting (ii) from (i), we get:
86x − 86y = 86
⇒ 86(x − y) = 86
⇒ x − y = 1 ...............(iv)

On adding (iii) and (iv), we get :
2x = 6 ⇒ x = 3

On substituting x = 3 in (iii), we get:
3 + y = 5
⇒ y = 5 − 3 = 2

Hence, the required solution is x = 3 and y = 2.

Page No 105:

Question 34:

Solve for x and y:
$23 x - 29 y = 98 29 x - 23 y = 110$

Answer:

The given equations are
$23 x - 29 y = 98 . . . . . (i) 29 x - 23 y = 110 . . . . . (i i)$
Adding (i) and (ii), we get
$52 x - 52 y = 208 \Rightarrow x - y = 4 . . . . . (i i i)$
Subtracting (i) from (ii), we get
$6 x + 6 y = 12 \Rightarrow x + y = 2 . . . . . (i v)$
Now, adding equation (iii) and (iv), we get
$2 x = 6 \Rightarrow x = 3$
Substituting x = 3 in (iv), we have
$3 + y = 2 \Rightarrow y = 2 - 3 = - 1$
Hence, $x = 3 and y = - 1$ .

Page No 105:

Question 35:

Solve for x and y:
$\frac{2 x + 5 y}{x y} = 6 \frac{4 x - 5 y}{x y} = - 3$

Answer:

The given equations can be written as
$\frac{5}{x} + \frac{2}{y} = 6 . . . . . (i) \frac{- 5}{x} + \frac{4}{y} = - 3 . . . . . (i i)$
Adding (i) and (ii), we get
$\frac{6}{y} = 3 \Rightarrow y = 2$
Substituting y = 2 in (i), we have
$\frac{5}{x} + \frac{2}{2} = 6 \Rightarrow x = 1$
Hence, x = 1 and y = 2..

Page No 105:

Question 36:

Solve for x and y:
$\frac{1}{(3 x + y)} + \frac{1}{(3 x - y)} = \frac{3}{4} \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}$

Answer:

The given equations are
$\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4} . . . . . (i) \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = - \frac{1}{8} \frac{1}{3 x + y} - \frac{1}{3 x - y} = - \frac{1}{4} (Multiplying by 2) . . . . . (i i)$
Substituting $\frac{1}{3 x + y} = u and \frac{1}{3 x - y} = v$ in (i) and (ii), we get
$u + v = \frac{3}{4} . . . . . (i i i) u - v = - \frac{1}{4} . . . . . (i v)$
Adding (iii) and (iv), we get
$2 u = \frac{1}{2} \Rightarrow u = \frac{1}{4} \Rightarrow 3 x + y = 4 (∵ \frac{1}{3 x + y} = u) . . . . . (v)$
Now, substituting $u = \frac{1}{4}$ in (iii), we get
$\frac{1}{4} + v = \frac{3}{4} v = \frac{3}{4} - \frac{1}{4} \Rightarrow v = \frac{1}{2} \Rightarrow 3 x - y = 2 (∵ \frac{1}{3 x - y} = v) . . . . . (v i)$
Adding (v) and (vi), we get
$6 x = 6 \Rightarrow x = 1$
Substituting x = 1 in (v), we have
$3 + y = 4 \Rightarrow y = 1$
Hence, x = 1 and y = 1.

Page No 105:

Question 37:

$\frac{1}{2 (x + 2 y)} + \frac{5}{3 (3 x - 2 y)} = - \frac{3}{2}, \frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60,} where x + 2 y \neq 0 and 3 x - 2 y \neq 0 .$

Answer:

The given equations are:
$\frac{1}{2 (x + 2 y)} + \frac{5}{3 (3 x - 2 y)} = - \frac{3}{2}$               ...(i)
$\frac{5}{4 (x + 2 y)} - \frac{3}{5 (3 x - 2 y)} = \frac{61}{60}$               ...(ii)
Putting $\frac{1}{x + 2 y} = u$ and $\frac{1}{3 x - 2 y} = v$ , we get:
$\frac{1}{2} u + \frac{5}{3} v = - \frac{3}{2}$          ...(iii)
$\frac{5}{4} u - \frac{3}{5} v = \frac{61}{60}$           ...(iv)
On multiplying (iii) by 6 and (iv) by 20, we get:
3u + 10v = −9         ...(v)
$25 u - 12 v = \frac{61}{3}$            ...(vi)
On multiplying (v) by 6 and (vi) by 5, we get:
18u + 60v = −54       ...(vii)
$125 u - 60 v = \frac{305}{3}$           ...(viii)
On adding (vii) and (viii), we get:
$143 u = \frac{305}{3} - 54 = \frac{305 - 162}{3} = \frac{143}{3}$
$\Rightarrow u = \frac{1}{3} = \frac{1}{x + 2 y}$
⇒ x + 2y = 3    ...(ix)
On substituting $u = \frac{1}{3}$ in (v), we get:
1 + 10v = −9
⇒ 10v = −10
⇒ v = −1
$\Rightarrow \frac{1}{3 x - 2 y} = - 1 \Rightarrow 3 x - 2 y = - 1$             ...(x)
On adding (ix) and (x), we get:
4x = 2
⇒ $x = \frac{1}{2}$
On substituting $x = \frac{1}{2}$ in (x), we get:
$\frac{3}{2} - 2 y = - 1 \Rightarrow 2 y = (\frac{3}{2} + 1) = \frac{5}{2} \Rightarrow y = \frac{5}{4}$
Hence, the required solution is $x = \frac{1}{2} and y = \frac{5}{4}$ .

Page No 105:

Question 38:

Solve for x and y:
$\frac{2}{(3 x + 2 y)} + \frac{3}{(3 x - 2 y)} = \frac{17}{5} \frac{5}{(3 x + 2 y)} + \frac{1}{(3 x - 2 y)} = 2$

Answer:

The given equations are
$\frac{2}{3 x + 2 y} + \frac{3}{3 x - 2 y} = \frac{17}{5} . . . . . (i) \frac{5}{3 x + 2 y} + \frac{1}{3 x - 2 y} = 2 . . . . . (i i)$
Substituting $\frac{1}{3 x + 2 y} = u and \frac{1}{3 x - 2 y} = v$ in (i) and (ii), we get
$2 u + 3 v = \frac{17}{5} . . . . . (i i i) 5 u + v = 2 . . . . . (i v)$
Multiplying (iv) by 3 and subtracting from(iii), we get
$2 u - 15 u = \frac{17}{5} - 6 \Rightarrow - 13 u = \frac{- 13}{5} \Rightarrow u = \frac{1}{5} \Rightarrow 3 x + 2 y = 5 (∵ \frac{1}{3 x + 2 y} = u) . . . . . (v)$
Now, substituting $u = \frac{1}{5}$ in (iv), we get
$1 + v = 2 \Rightarrow v = 1 \Rightarrow 3 x - 2 y = 1 (∵ \frac{1}{3 x - 2 y} = v) . . . . . (v i)$
Adding (v) and (vi), we get
$6 x = 6 \Rightarrow x = 1$
Substituting x = 1 in (v), we have
$3 + 2 y = 5 \Rightarrow y = 1$
Hence, x = 1 and y = 1.

Page No 105:

Question 39:

Solve for x and y:
$3 (2 x + y) = 7 x y 3 (x + 3 y) = 11 x y$

Answer:

The given equations can be written as
$\frac{3}{x} + \frac{6}{y} = 7 . . . . . (i) \frac{9}{x} + \frac{3}{y} = 11 . . . . . (i i)$
Multiplying (i) by 3 and subtracting (ii) from it, we get
$\frac{18}{y} - \frac{3}{y} = 21 - 11 \Rightarrow \frac{15}{y} = 10 \Rightarrow y = \frac{15}{10} = \frac{3}{2}$
Substituting $y = \frac{3}{2}$ in (i), we have
$\frac{3}{x} + \frac{6 \times 2}{3} = 7 \Rightarrow \frac{3}{x} = 7 - 4 = 3 \Rightarrow x = 1$
Hence, $x = 1 and y = \frac{3}{2}$ .

Page No 105:

Question 40:

Solve for x and y:
$x + y = a + b a x - b y = a^{2} - b^{2}$

Answer:

The given equations are
$x + y = a + b . . . . . (i) a x - b y = a^{2} - b^{2} . . . . . (i i)$
Multiplying (i) by b and adding it with (ii), we get
$b x + a x = a b + b^{2} + a^{2} - b^{2} \Rightarrow x = \frac{a b + a^{2}}{a + b} = a$
Substituting x = a in (i), we have
$a + y = a + b \Rightarrow y = b$
Hence, x = a and y = b.

Page No 105:

Question 41:

$\frac{x}{a} + \frac{y}{b} = 2, a x - b y = (a^{2} - b^{2}) .$

Answer:

The given equations are:
$\frac{x}{a} + \frac{y}{b} = 2$

⇒ $\frac{b x + a y}{a b} = 2$ [Taking LCM]
⇒bx + ay = 2ab .......(i)

Again, ax − by = (a² − b²) ........(ii)

On multiplying (i) by b and (ii) by a, we get:
b²x + bay = 2ab².........(iii)
a²x − bay = a(a² − b²) ........(iv)

On adding (iii) from (iv), we get:
(b² + a²)x = 2a²b + a(a² − b²)
⇒ (b² + a²)x = 2ab² + a³ − ab²
⇒ (b² + a²)x = ab²+ a³
⇒ (b² + a²)x = a(b² + a²)
⇒ $x = \frac{a (b^{2} + a^{2})}{(b^{2} + a^{2})} = a$
On substituting x = a in (i), we get:
ba + ay = 2ab
⇒ ay = ab
⇒ y = b

Hence, the solution is x = a and y = b.

Page No 105:

Question 42:

Solve for x and y:
$p x + q y = p - q q x - p y = p + q$

Answer:

The given equations are
$p x + q y = p - q . . . . . (i) q x - p y = p + q . . . . . (i i)$
Multiplying (i) by p and (ii) by q and adding them, we get
$p^{2} x + q^{2} x = p^{2} - p q + p q + q^{2} \Rightarrow x = \frac{p^{2} + q^{2}}{p^{2} + q^{2}} = 1$
Substituting x = 1 in (i), we have
$p + q y = p - q \Rightarrow q y = - q \Rightarrow y = - 1$
Hence, x = 1 and $y = - 1$ .

Page No 105:

Question 43:

Solve for x and y:
$\frac{x}{a} - \frac{y}{b} = 0 a x + b y = a^{2} + b^{2}$

Answer:

The given equations are
$\frac{x}{a} - \frac{y}{b} = 0 . . . . . (i) a x + b y = a^{2} + b^{2} . . . . . (i i)$
From (i)
$y = \frac{b x}{a}$
Substituting $y = \frac{b x}{a}$ in (ii), we get
$a x + \frac{b \times b x}{a} = a^{2} + b^{2} \Rightarrow x = \frac{(a^{2} + b^{2}) \times a}{a^{2} + b^{2}} = a$
Now, substitute x = a in (ii) to get
$a^{2} + b y = a^{2} + b^{2} \Rightarrow b y = b^{2} \Rightarrow y = b$
Hence, x = a and y = b.

Page No 105:

Question 44:

$6 (a x + b y) = 3 a + 2 b, 6 (b x - a y) = 3 b - 2 a .$

Answer:

The given equations are:
6(ax + by) = 3a + 2b
⇒ 6ax + 6by = 3a + 2b ...............(i)

and 6(bx − ay) = 3b − 2a
⇒ 6bx − 6ay = 3b − 2a ...................(ii)

On multiplying (i) by a and (ii) by b, we get:
6a²x + 6aby = 3a² + 2ab ................(iii)
6b²x − 6aby = 3b² − 2ab ....................(iv)

On adding (iii) and (iv), we get:
6(a² + b²)x = 3(a² + b²)
$\Rightarrow x = \frac{3 (a^{2} + b^{2})}{6 (a^{2} + b^{2})} = \frac{1}{2}$

On substituting $x = \frac{1}{2}$ in (i), we get:
$6 a \times \frac{1}{2} + 6 b y = 3 a + 2 b$
⇒ 3a + 6by = 3a + 2b
⇒ 6by = 2b
⇒ y = $\frac{2 b}{6 b} = \frac{1}{3}$
Hence, the required solution is $x = \frac{1}{2}$ and $y = \frac{1}{3}$ .

Page No 105:

Question 45:

Solve for x and y:
$a x - b y = a^{2} + b^{2} x + y = 2 a$

Answer:

The given equations are
$a x - b y = a^{2} + b^{2} . . . . . (i) x + y = 2 a . . . . . (i i)$
From (ii)
$y = 2 a - x$
Substituting $y = 2 a - x$ in (i), we get
$a x - b (2 a - x) = a^{2} + b^{2} \Rightarrow a x - 2 a b + b x = a^{2} + b^{2} \Rightarrow x = \frac{a^{2} + b^{2} + 2 a b}{a + b} = \frac{{(a + b)}^{2}}{a + b} = a + b$
Now, substitute x = a + b in (ii) to get
$a + b + y = 2 a \Rightarrow y = a - b$
Hence, $x = a + b and y = a - b$ .

Page No 105:

Question 46:

$\frac{b x}{a} - \frac{a y}{b} + a + b = 0, b x - a y + 2 a b = 0 .$

Answer:

The given equations are:
$\frac{b x}{a} - \frac{a y}{b} + a + b = 0$
By taking LCM, we get:
b²x − a²y = −a²b − b²a .......(i)

and bx − ay + 2ab = 0
bx − ay = −2ab ........(ii)

On multiplying (ii) by a, we get:
abx − a²y = −2a²b .......(iii)

On subtracting (i) from (iii), we get:
abx − b²x = − 2a²b + a²b + b²a = −a²b + b²a
⇒ x(ab − b²) = −ab(a − b)
⇒ x(a − b)b = −ab(a − b)
∴ $x = \frac{- a b (a - b)}{(a - b) b} = - a$

On substituting x = −a in (i), we get:
b²(−a) − a²y = −a²b − b²a
⇒ −b²a − a²y = −a²b − b²a
⇒ −a²y = −a²b
⇒ y = b

Hence, the solution is x = −a and y = b.

Page No 105:

Question 47:

$\frac{b x}{a} + \frac{a y}{b} = a^{2} + b^{2}, x + y = 2 a b .$

Answer:

The given equations are:

$\frac{b x}{a} + \frac{a y}{b} = a^{2} + b^{2}$
By taking LCM, we get:
$\frac{b^{2} x + a^{2} y}{a b} = a^{2} + b^{2}$
⇒ b²x + a²y = (ab)a² + b²
⇒ b²x + a²y = a³b + ab³ .......(i)

Also, x + y = 2ab........(ii)

On multiplying (ii) by a², we get:
a²x + a²y = 2a³b.........(iii)

On subtracting (iii) from (i), we get:
(b² − a²)x = a³b + ab³ − 2a³b
⇒ (b² − a²)x = −a³b + ab³
⇒ (b² − a²)x = ab(b² − a²)
⇒ (b² − a²)x = ab(b² − a²)
⇒ $x = \frac{a b (b^{2} - a^{2})}{(b^{2} - a^{2})} = a b$

On substituting x = ab in (i), we get:
b²(ab) + a²y = a³b + ab³
⇒ a²y = a³b
⇒ $\frac{a^{3} b}{a^{2}} = a b$
Hence, the solution is x = ab and y = ab.

Page No 105:

Question 48:

Solve for x and y:
$x + y = a + b a x - b y = a^{2} - b^{2}$

Answer:

The given equations are
$x + y = a + b . . . . . (i) a x - b y = a^{2} - b^{2} . . . . . (i i)$
From (i)
$y = a + b - x$
Substituting $y = a + b - x$ in (ii), we get
$a x - b (a + b - x) = a^{2} - b^{2} \Rightarrow a x - a b - b^{2} + b x = a^{2} - b^{2} \Rightarrow x = \frac{a^{2} + a b}{a + b} = a$
Now, substitute x = a in (i) to get
$a + y = a + b \Rightarrow y = b$
Hence, x = a and y = b.

Page No 105:

Question 49:

Solve for x and y:
$a^{2} x + b^{2} y = c^{2} b^{2} x + a^{2} y = d^{2}$

Answer:

The given equations are
$a^{2} x + b^{2} y = c^{2} . . . . . (i) b^{2} x + a^{2} y = d^{2} . . . . . (i i)$
Multiplying (i) by a²and (ii) by b² and subtracting, we get
$a^{4} x - b^{4} x = a^{2} c^{2} - b^{2} d^{2} \Rightarrow x = \frac{a^{2} c^{2} - b^{2} d^{2}}{a^{4} - b^{4}}$
Now, multiplying (i) by b²and (ii) by a² and subtracting, we get
$b^{4} y - a^{4} y = b^{2} c^{2} - a^{2} d^{2} \Rightarrow y = \frac{b^{2} c^{2} - a^{2} d^{2}}{b^{4} - a^{4}}$

Hence, $x = \frac{a^{2} c^{2} - b^{2} d^{2}}{a^{4} - b^{4}} and y = \frac{b^{2} c^{2} - a^{2} d^{2}}{b^{4} - a^{4}}$ .

Page No 105:

Question 50:

Solve for x and y:
$\frac{x}{a} + \frac{y}{b} = a + b \frac{x}{a^{2}} + \frac{y}{b^{2}} = 2$

Answer:

The given equations are
$\frac{x}{a} + \frac{y}{b} = a + b . . . . . (i) \frac{x}{a^{2}} + \frac{y}{b^{2}} = 2 . . . . . (i i)$
Multiplying (i) by band (ii) by b² and subtracting, we get
$\frac{b x}{a} - \frac{b^{2} x}{a^{2}} = a b + b^{2} - 2 b^{2} \Rightarrow \frac{a b - b^{2}}{a^{2}} x = a b - b^{2} \Rightarrow x = \frac{(a b - b^{2}) a^{2}}{a b - b^{2}} = a^{2}$
Now, substituting x = a²in (i), we get
$\frac{a^{2}}{a} + \frac{y}{b} = a + b \Rightarrow \frac{y}{b} = a + b - a = b \Rightarrow y = b^{2}$

Hence, $x = a^{2} and y = b^{2}$ .

Page No 111:

Question 1:

$x + 2 y + 1 = 0, 2 x - 3 y - 12 = 0 .$

Answer:

The given equations are:
x + 2y + 1 = 0 ...(i)
2x − 3y − 12 = 0 ...(ii)
Here, a₁ = 1, b₁= 2, c₁ = 1, a₂ = 2, b₂ = −3 and c₂ = −12
By cross multiplication, we have:

∴ $\frac{x}{[2 \times (- 12) - 1 \times (- 3)]} = \frac{y}{[1 \times 2 - 1 \times (- 12)]} = \frac{1}{[1 \times (- 3) - 2 \times 2]}$
⇒ $\frac{x}{(- 24 + 3)} = \frac{y}{(2 + 12)} = \frac{1}{(- 3 - 4)}$
⇒ $\frac{x}{(- 21)} = \frac{y}{(14)} = \frac{1}{(- 7)}$
⇒ $x = \frac{- 21}{- 7} = 3, y = \frac{14}{- 7} = - 2$
Hence, x = 3 and y = −2 is the required solution.

Page No 111:

Question 2:

$3 x - 2 y + 3 = 0, 4 x + 3 y - 47 = 0 .$

Answer:

The given equations are:
3x − 2y + 3 = 0 ...(i)
4x + 3y − 47 = 0 ...(ii)
Here, a₁ = 3, b₁= −2 , c₁ = 3, a₂ = 4, b₂ = 3 and c₂ = −47
By cross multiplication, we have:

∴ $\frac{x}{[(- 2) \times (- 47) - 3 \times 3]} = \frac{y}{[3 \times 4 - (- 47) \times 3]} = \frac{1}{[3 \times 3 - (- 2) \times 4]}$
⇒ $\frac{x}{(94 - 9)} = \frac{y}{(12 + 141)} = \frac{z}{(9 + 8)}$
⇒ $\frac{x}{85} = \frac{y}{153} = \frac{1}{17}$
⇒ $x = \frac{85}{17} = 5, y = \frac{153}{17} = 9$
Hence, x = 5 and y = 9 is the required solution.

Page No 111:

Question 3:

$6 x - 5 y - 16 = 0, 7 x - 13 y + 10 = 0 .$

Answer:

The given equations are:
6x − 5y − 16 = 0 ...(i)
7x − 13y + 10 = 0 ...(ii)
Here, a₁ = 6, b₁= −5 , c₁ = −16, a₂ = 7, b₂ = −13 and c₂ = 10
By cross multiplication, we have:

∴ $\frac{x}{[(- 5) \times 10 - (- 16) \times (- 13)]} = \frac{y}{[(- 16) \times 7 - 10 \times 6]} = \frac{1}{[6 \times (- 13) - (- 5) \times 7]}$
⇒ $\frac{x}{(- 50 - 208)} = \frac{y}{(- 112 - 60)} = \frac{z}{(- 78 + 35)}$
⇒ $\frac{x}{(- 258)} = \frac{y}{(- 172)} = \frac{1}{(- 43)}$
⇒ $x = \frac{- 258}{- 43} = 6, y = \frac{- 172}{- 43} = 4$
Hence, x = 6 and y = 4 is the required solution.

Page No 111:

Question 4:

$3 x + 2 y + 25 = 0, 2 x + y + 10 = 0 .$

Answer:

The given equations are:
3x + 2y + 25 = 0 ...(i)
2x + y + 10 = 0 ...(ii)
Here, a₁ = 3, b₁= 2 , c₁ = 25, a₂ = 2, b₂ = 1 and c₂ = 10
By cross multiplication, we have:

∴ $\frac{x}{[2 \times 10 - 25 \times 1]} = \frac{y}{[25 \times 2 - 10 \times 3]} = \frac{1}{[3 \times 1 - 2 \times 2]}$
⇒ $\frac{x}{(20 - 25)} = \frac{y}{(50 - 30)} = \frac{1}{(3 - 4)}$
⇒ $\frac{x}{(- 5)} = \frac{y}{20} = \frac{1}{(- 1)}$
⇒ $x = \frac{- 5}{- 1} = 5, y = \frac{20}{(- 1)} = - 20$
Hence, x = 5 and y = −20 is the required solution.

Page No 111:

Question 5:

$2 x + 5 y = 1, 2 x + 3 y = 3 .$

Answer:

The given equations may be written as:
2x + 5y − 1 = 0 ...(i)
2x + 3y − 3 = 0 ...(ii)
Here, a₁ = 2, b₁= 5, c₁ = −1, a₂ = 2, b₂ = 3 and c₂ = −3
By cross multiplication, we have:

∴ $\frac{x}{[5 \times (- 3) - 3 \times (- 1)]} = \frac{y}{[(- 1) \times 2 - (- 3) \times 2]} = \frac{1}{[2 \times 3 - 2 \times 5]}$
⇒ $\frac{x}{(- 15 + 3)} = \frac{y}{(- 2 + 6)} = \frac{z}{(6 - 10)}$
⇒ $\frac{x}{- 12} = \frac{y}{4} = \frac{1}{- 4}$
⇒ $x = \frac{- 12}{- 4} = 3, y = \frac{4}{- 4} = - 1$
Hence, x = 3 and y = −1 is the required solution.

Page No 111:

Question 6:

$2 x + y = 35, 3 x + 4 y = 65 .$

Answer:

The given equations may be written as:
2x + y − 35 = 0 ...(i)
3x + 4y − 65 = 0 ...(ii)
Here, a₁ = 2, b₁= 1, c₁ = −35, a₂ = 3, b₂ = 4 and c₂ = −65
By cross multiplication, we have:

∴ $\frac{x}{[1 \times (- 65) - 4 \times (- 35)]} = \frac{y}{[(- 35) \times 3 - (- 65) \times 2]} = \frac{1}{[2 \times 4 - 3 \times 1]}$
⇒ $\frac{x}{(- 65 + 140)} = \frac{y}{(- 105 + 130)} = \frac{1}{(8 - 3)}$
⇒ $\frac{x}{75} = \frac{y}{25} = \frac{1}{5}$
⇒ $x = \frac{75}{5} = 15, y = \frac{25}{5} = 5$
Hence, x = 15 and y = 5 is the required solution.

Page No 111:

Question 7:

$7 x - 2 y = 3, 22 x - 3 y = 16 .$

Answer:

The given equations may be written as:
7x − 2y − 3 = 0 ...(i)
$22 x - 3 y - 16 = 0$ ...(ii)
Here, a₁ = 7, b₁= −2 , c₁ = −3, a₂ = 22, b₂ = $- 3$ and c₂ = −16
By cross multiplication, we have:

∴ $\frac{x}{[(- 2) \times (- 16) - (- 3) \times (- 3)]} = \frac{y}{[(- 3) \times 22 - (- 16) \times 7]} = \frac{1}{[7 \times (- 3) - 22 \times (- 2)]}$
⇒ $\frac{x}{(32 - 9)} = \frac{y}{(- 66 + 112)} = \frac{1}{(- 21 + 44)}$
⇒ $\frac{x}{(23)} = \frac{y}{46} = \frac{1}{(23)}$
⇒ $x = \frac{23}{23} = 1, y = \frac{46}{23} = 2$
Hence, x = 1 and y = 2 is the required solution.

Page No 111:

Question 8:

$\frac{x}{6} + \frac{y}{15} = 4, \frac{x}{3} - \frac{y}{12} = \frac{19}{4} .$

Answer:

The given equations may be written as:
$\frac{x}{6} + \frac{y}{15} - 4 = 0$ ...(i)
$\frac{x}{3} - \frac{y}{12} - \frac{19}{4} = 0$ ...(ii)
Here, $a_{1} = \frac{1}{6}, b_{1} = \frac{1}{15}, c_{1} = - 4, a_{2} = \frac{1}{3}, b_{2} = - \frac{1}{12} and c_{2} = - \frac{19}{4}$
By cross multiplication, we have:

∴ $\frac{x}{[\frac{1}{15} \times (- \frac{19}{4}) - (- \frac{1}{12}) \times (- 4)]} = \frac{y}{[(- 4) \times \frac{1}{3} - (\frac{1}{6}) \times (- \frac{19}{4})]} = \frac{1}{[\frac{1}{6} \times (- \frac{1}{12}) - \frac{1}{3} \times \frac{1}{15}]}$
⇒ $\frac{x}{(- \frac{19}{60} - \frac{1}{3})} = \frac{y}{(- \frac{4}{3} + \frac{19}{24})} = \frac{1}{(- \frac{1}{72} - \frac{1}{45})}$
⇒ $\frac{x}{(- \frac{39}{60})} = \frac{y}{(- \frac{13}{24})} = \frac{1}{(- \frac{13}{360})}$
⇒ $x = [(- \frac{39}{60}) \times (- \frac{360}{13})] = 18, y = [(- \frac{13}{24}) \times (- \frac{360}{13})] = 15$
Hence, x = 18 and y = 15 is the required solution.

Page No 111:

Question 9:

$\frac{1}{x} + \frac{1}{y} = 7, \frac{2}{x} + \frac{3}{y} = 17 (x \neq 0, y \neq 0) .$

Answer:

Taking $\frac{1}{x} = u$ and $\frac{1}{y} = v$ , the given equations become:
u + v = 7
2u + 3v = 17

The given equations may be written as:
u + v − 7 = 0            ...(i)
2u + 3v − 17 = 0      ...(ii)

Here, a₁= 1, b₁ = 1, c₁ = −7, a₂ = 2, b₂ = 3 and c₂ = −17
By cross multiplication, we have:

∴ $\frac{u}{[1 \times (- 17) - 3 \times (- 7)]} = \frac{v}{[(- 7) \times 2 - 1 \times (- 17)]} = \frac{1}{[3 - 2]}$
⇒ $\frac{u}{- 17 + 21} = \frac{v}{- 14 + 17} = \frac{1}{1}$
⇒ $\frac{u}{4} = \frac{v}{3} = \frac{1}{1}$
⇒ $u = \frac{4}{1} = 4, v = \frac{3}{1} = 3$
⇒ $\frac{1}{x} = 4, \frac{1}{y} = 3$
⇒ $x = \frac{1}{4}, y = \frac{1}{3}$
Hence, $x = \frac{1}{4}$ and $y = \frac{1}{3}$ is the required solution.

Page No 111:

Question 10:

$\frac{5}{(x + y)} - \frac{2}{(x - y)} + 1 = 0, \frac{15}{(x + y)} + \frac{7}{(x - y)} - 10 = 0 (x \neq y, x \neq - y) .$

Answer:

Taking $\frac{1}{x + y} = u$ and $\frac{1}{x - y} = v$ , the given equations become:
5u − 2v + 1 = 0      ...(i)
15u + 7v − 10 = 0    ...(ii)
Here, a₁= 5, b₁ = −2, c₁ = 1, a₂ = 15, b₂ = −7 and c₂ = −10
By cross multiplication, we have:

∴ $\frac{u}{[- 2 \times (- 10) - 1 \times 7]} = \frac{v}{[1 \times 15 - (- 10) \times 5]} = \frac{1}{[35 + 30]}$
⇒ $\frac{u}{20 - 7} = \frac{v}{15 + 50} = \frac{1}{65}$
⇒ $\frac{u}{13} = \frac{v}{65} = \frac{1}{65}$
⇒ $u = \frac{13}{65} = \frac{1}{5}, v = \frac{65}{65} = 1$
⇒ $\frac{1}{x + y} = \frac{1}{5}, \frac{1}{x - y} = 1$
So, (x + y) = 5            ...(iii)
and (x − y) = 1           ...(iv)

Again, the above equations (iii) and (iv) may be written as:
x + y − 5 = 0           ...(v)
x − y − 1 = 0          ...(vi)
Here, a₁= 1, b₁ = 1, c₁ = −5, a₂ = 1, b₂ = −1 and c₂ = −1
By cross multiplication, we have:

∴ $\frac{x}{[1 \times (- 1) - (- 5) \times (- 1)]} = \frac{y}{[(- 5) \times 1 - (- 1) \times 1]} = \frac{1}{[1 \times (- 1) - 1 \times 1]}$

⇒ $\frac{x}{(- 1 - 5)} = \frac{y}{(- 5 + 1)} = \frac{1}{(- 1 - 1)}$
⇒ $\frac{x}{- 6} = \frac{y}{- 4} = \frac{1}{- 2}$
⇒ $x = \frac{- 6}{- 2} = 3, y = \frac{- 4}{- 2} = 2$
Hence, x = 3 and y = 2 is the required solution.

Page No 111:

Question 11:

$\frac{a x}{b} - \frac{b y}{a} = a + b, a x - b y = 2 a b$

Answer:

The given equations may be written as:
$\frac{a x}{b} - \frac{b y}{a} - (a + b) = 0$ ...(i)
$a x - b y - 2 a b = 0$ ...(ii)
Here, a₁= $\frac{a}{b}$ , b₁ = $\frac{- b}{a}$ , c₁ = −(a + b), a₂ = a, b₂ = −b and c₂ = −2ab
By cross multiplication, we have:

∴ $\frac{x}{(- \frac{b}{a}) \times (- 2 a b) - (- b) \times (- (a + b))} = \frac{y}{- (a + b) \times a - (- 2 a b) \times \frac{a}{b}} = \frac{1}{\frac{a}{b} \times (- b) - a \times (- \frac{b}{a})}$
⇒ $\frac{x}{2 b^{2} - b (a + b)} = \frac{y}{- a (a + b) + 2 a^{2}} = \frac{1}{- a + b}$
⇒ $\frac{x}{2 b^{2} - a b - b^{2}} = \frac{y}{- a^{2} - a b + 2 a^{2}} = \frac{1}{- a + b}$
⇒ $\frac{x}{b^{2} - a b} = \frac{y}{a^{2} - a b} = \frac{1}{- (a - b)}$
⇒ $\frac{x}{- b (a - b)} = \frac{y}{a (a - b)} = \frac{1}{- (a - b)}$
⇒ $x = \frac{- b (a - b)}{- (a - b)} = b, y = \frac{a (a - b)}{- (a - b)} = - a$
Hence, x = b and y = −a is the required solution.

Page No 111:

Question 12:

$2 a x + 3 b y = (a + 2 b), 3 a x + 2 b y = (2 a + b) .$

Answer:

The given equations may be written as:
2ax + 3by − (a + 2b) = 0 ...(i)
3ax + 2by − (2a + b) = 0 ...(ii)
Here, a₁= 2a, b₁ = 3b, c₁ = −(a + 2b), a₂ = 3a, b₂ = 2b and c₂ = −(2a + b)
By cross multiplication, we have:

∴ $\frac{x}{[3 b \times (- (2 a + b)) - 2 b \times (- (a + 2 b))]} = \frac{y}{[- (a + 2 b) \times 3 a - 2 a \times (- (2 a + b))]} = \frac{1}{[2 a \times 2 b - 3 a \times 3 b]}$
⇒ $\frac{x}{(- 6 a b - 3 b^{2} + 2 a b + 4 b^{2})} = \frac{y}{(- 3 a^{2} - 6 a b + 4 a^{2} + 2 a b)} = \frac{1}{4 a b - 9 a b}$
⇒ $\frac{x}{b^{2} - 4 a b} = \frac{y}{a^{2} - 4 a b} = \frac{1}{- 5 a b}$
⇒ $\frac{x}{- b (4 a - b)} = \frac{y}{- a (4 b - a)} = \frac{1}{- 5 a b}$
⇒ $x = \frac{- b (4 a - b)}{- 5 a b} = \frac{(4 a - b)}{5 a}, y = \frac{- a (4 b - a)}{- 5 a b} = \frac{(4 b - a)}{5 b}$
Hence, $x = \frac{(4 a - b)}{5 a}$ and $y = \frac{(4 b - a)}{5 b}$ is the required solution.

Page No 111:

Question 13:

Solve the following system of equations by using the method of cross-multiplication:
$\frac{a}{x} - \frac{b}{y} = 0 \frac{a b^{2}}{x} + \frac{a^{2} b}{y} = a^{2} + b^{2}, where x \neq 0 and y \neq 0$

Answer:

Substituting $\frac{1}{x} = u and \frac{1}{y} = v$ in the given equations, we get
$a u - b v + 0 = 0 . . . . . (i) a b^{2} u + a^{2} b v - (a^{2} + b^{2}) = 0 . . . . . (i i)$
Here, $a_{1} = a, b_{1} = - b, c_{1} = 0 and a_{2} = a b^{2}, b_{2} = a^{2} b, c_{2} = - (a^{2} + b^{2})$ .
So, by cross-multiplication, we have
$\frac{u}{b_{1} c_{2} - b_{2} c_{1}} = \frac{v}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}} \Rightarrow \frac{u}{(- b) [- (a^{2} + b^{2})] - (a^{2} b) (0)} = \frac{v}{(0) (a b^{2}) - (- a^{2} - b^{2}) (a)} = \frac{1}{(a) (a^{2} b) - (a b^{2}) (- b)} \Rightarrow \frac{u}{b (a^{2} + b^{2})} = \frac{v}{a (a^{2} + b^{2})} = \frac{1}{a b (a^{2} + b^{2})}$
$\Rightarrow u = \frac{b (a^{2} + b^{2})}{a b (a^{2} + b^{2})}, v = \frac{a (a^{2} + b^{2})}{a b (a^{2} + b^{2})} \Rightarrow u = \frac{1}{a}, v = \frac{1}{b} \Rightarrow \frac{1}{x} = \frac{1}{a}, \frac{1}{y} = \frac{1}{b} \Rightarrow x = a, y = b$
Hence, x = a and y = b.

Page No 122:

Question 1:

Show that the following system of equations has a unique solution:
$3 x + 5 y = 12, 5 x + 3 y = 4 .$
Also, find the solution of the given system of equations.

Answer:

The given system of equations is:
3x + 5y = 12
5x + 3y = 4
These equations are of the forms:
a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 3, b₁= 5, c₁ = −12 and a₂ = 5, b₂= 3, c₂ = −4
For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e., $\frac{3}{5} \neq \frac{5}{3}$
Hence, the given system of equations has a unique solution.

Again, the given equations are:
3x + 5y = 12            ...(i)
5x + 3y = 4              ...(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x + 15y = 36          ...(iii)
25x + 15y = 20        ...(iv)
On subtracting (iii) from (iv), we get:
16x = −16
⇒ x = −1
On substituting x = −1 in (i), we get:
3(−1) + 5y = 12
⇒ 5y = (12 + 3) = 15
⇒ y = 3
Hence, x = −1 and y = 3 is the required solution.

Page No 122:

Question 2:

Show that the following system of equations has a unique solution and solve it:
$2 x - 3 y = 17 4 x + y = 13$

Answer:

The system of equations can be written as
$2 x - 3 y - 17 = 0 . . . . . (i) 4 x + y - 13 = 0 . . . . . (i i)$
The given equations are of the form
$a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 2, b_{1} = - 3, c_{1} = - 17 and a_{2} = 4, b_{2} = 1, c_{2} = - 13$
Now,
$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2} and \frac{b_{1}}{b_{2}} = \frac{- 3}{1} = - 3$
Since, $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , therefore the system of equations has unique solution.
Using cross multiplication method, we have
$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}} \Rightarrow \frac{x}{- 3 (- 13) - 1 \times (- 17)} = \frac{y}{- 17 \times 4 - (- 13) \times 2} = \frac{1}{2 \times 1 - 4 \times (- 3)} \Rightarrow \frac{x}{39 + 17} = \frac{y}{- 68 + 26} = \frac{1}{2 + 12} \Rightarrow \frac{x}{56} = \frac{y}{- 42} = \frac{1}{14}$
$\Rightarrow x = \frac{56}{14}, y = \frac{- 42}{14} \Rightarrow x = 4, y = - 3$
Hence, $x = 4 and y = - 3$ .

Page No 122:

Question 3:

Show that the following system of equations has a unique solution:
$\frac{x}{3} + \frac{y}{2} = 3, x - 2 y = 2 .$
Also, find the solution of the given system of equations.

Answer:

The given system of equations are:
$\frac{x}{3} + \frac{y}{2} = 3$
⇒ $\frac{2 x + 3 y}{6} = 3$
2x + 3y = 18
⇒ 2x + 3y − 18 = 0                 ...(i)
and
x − 2y = 2
x − 2y − 2 = 0                    ...(ii)
These equations are of the forms:
a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 2, b₁= 3, c₁ = −18 and a₂ = 1, b₂= −2, c₂ = −2
For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e., $\frac{2}{1} \neq \frac{3}{- 2}$
Hence, the given system of equations has a unique solution.

Again, the given equations are:
2x + 3y − 18 = 0           ...(iii)
x − 2y − 2 = 0               ...(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y − 36 = 0           ...(v)
3x − 6y − 6 = 0             ...(vi)
On adding (v) and (vi), we get:
7x = 42
⇒ x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒ 3y = (18 − 12) = 6
⇒ y = 2
Hence, x = 6 and y = 2 is the required solution.

Page No 122:

Question 4:

Find the value of k for which the following equations has a unique solution:
$2 x + 3 y - 5 = 0 k x - 6 y - 8 = 0$

Answer:

The given system of equations are
$2 x + 3 y - 5 = 0 . . . . . (i) k x - 6 y - 8 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 2, b_{1} = 3, c_{1} = - 5 and a_{2} = k, b_{2} = - 6, c_{2} = - 8$
Now, for the given system of equations to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{2}{k} \neq \frac{3}{- 6} \Rightarrow k \neq - 4$
Hence, $k \neq - 4$ .

Page No 122:

Question 5:

Find the value of k for which the following equations has a unique solution:
$x - k y - 2 = 0 3 x + 2 y + 5 = 0$

Answer:

The given system of equations are
$x - k y - 2 = 0 . . . . . (i) 3 x + 2 y + 5 = 0 . . . . . (i i)$
This system of equations is of the form
$a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 1, b_{1} = - k, c_{1} = - 2 and a_{2} = 3, b_{2} = 2, c_{2} = 5$
Now, for the given system of equations to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{3} \neq \frac{- k}{2} \Rightarrow k \neq - \frac{2}{3}$
Hence, $k \neq - \frac{2}{3}$ .

Page No 122:

Question 6:

Find the value of k for which the following system of equations has a unique solution:
$5 x - 7 y - 5 = 0 2 x + k y - 1 = 0$

Answer:

The given system of equations is
$5 x - 7 y - 5 = 0 . . . . . (i) 2 x + k y - 1 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 5, b_{1} = - 7, c_{1} = - 5 and a_{2} = 2, b_{2} = k, c_{2} = - 1$
For the given system of equations to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{5}{2} \neq \frac{- 7}{k} \Rightarrow k \neq \frac{- 14}{5}$
Hence, $k \neq - \frac{14}{5}$ .

Page No 122:

Question 7:

Find the value of k for which the following system of equations has a unique solution:
$4 x + k y + 8 = 0 x + y + 1 = 0$

Answer:

The given system of equations is
$4 x + k y + 8 = 0 . . . . . (i) x + y + 1 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 4, b_{1} = k, c_{1} = 8 and a_{2} = 1, b_{2} = 1, c_{2} = 1$
For the given system of equations to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{4}{1} \neq \frac{k}{1} \Rightarrow k \neq 4$
Hence, $k \neq 4$ .

Page No 123:

Question 8:

Find the value of k for which each of the following systems of equations has a unique solution:
$4 x - 5 y = k, 2 x - 3 y = 12 .$

Answer:

The given system of equations:
4x − 5y = k
⇒ 4x − 5y − k = 0 ...(i)
And, 2x − 3y = 12
⇒ 2x − 3y − 12 = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 4, b₁= −5, c₁ = −k and a₂ = 2, b₂= −3, c₂ = −12
For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
i.e. $\frac{4}{2} \neq \frac{- 5}{- 3}$
$\Rightarrow 2 \neq \frac{5}{3} \Rightarrow 6 \neq 5$
Thus, for all real values of k, the given system of equations will have a unique solution.

Page No 123:

Question 9:

Find the value of k for which each of the following systems of equations has a unique solution:
$k x + 3 y = (k - 3), 12 x + k y = k .$

Answer:

The given system of equations:
kx + 3y = (k − 3)
⇒ kx + 3y − (k − 3) = 0 ....(i)
And, 12x + ky = k
⇒ 12x + ky − k = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 3, c₁ = −(k − 3) and a₂ = 12, b₂= k, c₂ = −k
For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
i.e. $\frac{k}{12} \neq \frac{3}{k}$
$\Rightarrow k^{2} \neq 36 \Rightarrow k \neq \pm 6$
Thus, for all real values of k other than $\pm 6$ , the given system of equations will have a unique solution.

Page No 123:

Question 10:

Show that the system of equations
$2 x - 3 y = 5, 6 x - 9 y = 15$
has an infinite number of solutions.

Answer:

The given system of equations is:
2x − 3y = 5
⇒ 2x − 3y − 5 = 0 ...(i)
6x − 9y = 15
⇒ 6x − 9y − 15 = 0 ...(ii)
These equations are of the forms:
a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 2, b₁= −3, c₁ = −5 and a₂ = 6, b₂= −9, c₂ = −15

$∴ \frac{a_{1}}{a_{2}} = \frac{2}{6} = \frac{1}{3}, \frac{b_{1}}{b 2} = \frac{- 3}{- 9} = \frac{1}{3}$ and $\frac{c_{1}}{c_{2}} = \frac{- 5}{- 15} = \frac{1}{3}$
Thus, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Hence, the given system of equations has an infinite number of solutions.

Page No 123:

Question 11:

Show that the system of equations $6 x + 5 y = 11, 9 x + \frac{15}{2} y = 21$ has no solution.

Answer:

The given system of equations can be written as
$6 x + 5 y - 11 = . . . . . (i) 9 x + \frac{15}{2} y - 21 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 6, b_{1} = 5, c_{1} = - 11 and a_{2} = 9, b_{2} = \frac{15}{2}, c_{2} = - 21$
Now,
$\frac{a_{1}}{a_{2}} = \frac{6}{9} = \frac{2}{3} \frac{b_{1}}{b_{2}} = \frac{5}{\frac{15}{2}} = \frac{2}{3} \frac{c_{1}}{c_{2}} = \frac{- 11}{- 21} = \frac{11}{21}$
Since, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ , therefore the given system has no solution.

Page No 123:

Question 12:

For what value of k, the system of equations
$k x + 2 y = 5, 3 x - 4 y = 10$
has (i) a unique solution, (ii) no solution?

Answer:

The given system of equations is:
kx + 2y = 5
⇒ kx + 2y − 5= 0 ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0 ...(ii)
These equations are of the forms:
a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0
where, a₁ = k, b₁= 2, c₁ = −5 and a₂ = 3, b₂= −4, c₂ = −10
(i) For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e., $\frac{k}{3} \neq \frac{2}{- 4} \Rightarrow k \neq \frac{- 3}{2}$
Thus for all real values of k other than $\frac{- 3}{2}$ , the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{k}{3} = \frac{2}{- 4} \neq \frac{- 5}{- 10}$
$\Rightarrow \frac{k}{3} = \frac{2}{- 4} and \frac{k}{3} \neq \frac{1}{2}$
$\Rightarrow k = \frac{- 3}{2}, k \neq \frac{3}{2}$
Hence, the required value of k is $\frac{- 3}{2}$ .

Page No 123:

Question 13:

For what value of k, the system of equations
$x + 2 y = 5, 3 x + k y + 15 = 0$
has (i) a unique solution, (ii) no solution?

Answer:

The given system of equations is:
x + 2y = 5
⇒ x + 2y − 5= 0 ...(i)
3x + ky + 15 = 0 ...(ii)
These equations are of the form:
a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 1, b₁= 2, c₁ = −5 and a₂ = 3, b₂= k, c₂ = 15
(i) For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e., $\frac{1}{3} \neq \frac{2}{k} \Rightarrow k \neq 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{1}{3} = \frac{2}{k} \neq \frac{- 5}{15}$
$\Rightarrow \frac{1}{3} = \frac{2}{k} and \frac{2}{k} \neq \frac{- 5}{15}$
$\Rightarrow k = 6, k \neq - 6$
Hence, the required value of k is 6.

Page No 123:

Question 14:

For what value of k does the system of equations
$x + 2 y = 3, 5 x + k y + 7 = 0$
has (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equations has infinitely many solutions.

Answer:

The given system of equations is:
x + 2y = 3
x + 2y − 3= 0 ....(i)
And, 5x + ky + 7 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁= 2, c₁ = −3 and a₂ = 5, b₂= k, c₂ = 7
(i) For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , i.e. $\frac{1}{5} \neq \frac{2}{k} \Rightarrow k \neq 10$
Thus, for all real values of k, other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{1}{5} = \frac{2}{k} \neq \frac{- 3}{7}$
$\Rightarrow \frac{1}{5} = \frac{2}{k} and \frac{2}{k} \neq \frac{- 3}{7}$
$\Rightarrow k = 10, k \neq \frac{14}{- 3}$
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.

Page No 123:

Question 15:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k .$

Answer:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0 ....(i)
And, (k − 1)x + (k + 2)y = 3k
⇒ (k − 1)x + (k + 2)y − 3k = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= 3, c₁ = −7 and a₂ = (k − 1), b₂= (k + 2), c₂ = −3k
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{2}{(k - 1)} = \frac{3}{(k + 2)} = \frac{- 7}{- 3 k}$
$\Rightarrow \frac{2}{(k - 1)} = \frac{3}{(k + 2)} = \frac{7}{3 k}$

Now, we have the following three cases:
Case I:
$\frac{2}{k - 1} = \frac{3}{k + 2}$
$\Rightarrow 2 (k + 2) = 3 (k - 1) \Rightarrow 2 k + 4 = 3 k - 3 \Rightarrow k = 7$

Case II:
$\frac{3}{k + 2} = \frac{7}{3 k}$
$\Rightarrow 7 (k + 2) = 9 k \Rightarrow 7 k + 14 = 9 k \Rightarrow 2 k = 14 \Rightarrow k = 7$

Case III:
$\frac{2}{k - 1} = \frac{7}{3 k}$
$\Rightarrow 7 k - 7 = 6 k \Rightarrow k = 7$

Hence, the given system of equations has an infinite number of solutions when k is equal to 7.

Page No 123:

Question 16:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$2 x + (k - 2) y = k, 6 + (2 k - 1) y = (2 k + 5) .$

Answer:

The given system of equations:
2x + (k − 2)y = k
⇒ 2x + (k − 2)y − k = 0 ....(i)
And, 6x + (2k − 1)y = (2k + 5)
⇒ 6x + (2k − 1)y − (2k + 5) = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= (k − 2), c₁ = −k and a₂ = 6, b₂= (2k − 1), c₂ = −(2k + 5)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{2}{6} = \frac{(k - 2)}{(2 k - 1)} = \frac{- k}{- (2 k + 5)}$
$\Rightarrow \frac{1}{3} = \frac{(k - 2)}{(2 k - 1)} = \frac{k}{(2 k + 5)}$

Now, we have the following three cases:
Case I:
$\frac{1}{3} = \frac{k - 2}{2 k - 1}$
$\Rightarrow (2 k - 1) = 3 (k - 2)$
$\Rightarrow 2 k - 1 = 3 k - 6 \Rightarrow k = 5$

Case II:
$\frac{k - 2}{2 k - 1} = \frac{k}{2 k + 5}$
$\Rightarrow (k - 2) (2 k + 5) = k (2 k - 1)$
$\Rightarrow 2 k^{2} + 5 k - 4 k - 10 = 2 k^{2} - k$
$\Rightarrow k + k = 10 \Rightarrow 2 k = 10 \Rightarrow k = 5$

Case III:
$\frac{1}{3} = \frac{k}{2 k + 5}$
$\Rightarrow 2 k + 5 = 3 k \Rightarrow k = 5$

Hence, the given system of equations has an infinite number of solutions when k is equal to 5.

Page No 123:

Question 17:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$k x + 3 y = (2 x + 1), 2 (k + 1) x + 9 y = (7 k + 1) .$

Answer:

The given system of equations:
kx + 3y = (2k + 1)
⇒ kx + 3y − (2k + 1) = 0 ...(i)
And, 2(k + 1)x + 9y = (7k + 1)
⇒ 2(k + 1)x + 9y − (7k + 1) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 3, c₁ = −(2k + 1) and a₂ = 2(k + 1), b₂= 9, c₂ = −(7k + 1)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$i . e . \frac{k}{2 (k + 1)} = \frac{3}{9} = \frac{- (2 k + 1)}{- (7 k + 1)}$
$\Rightarrow \frac{k}{2 (k + 1)} = \frac{1}{3} = \frac{(2 k + 1)}{(7 k + 1)}$

Now, we have the following three cases:
Case I:
$\frac{k}{2 (k + 1)} = \frac{1}{3}$
$\Rightarrow 2 (k + 1) = 3 k$
$\Rightarrow 2 k + 2 = 3 k$
$\Rightarrow k = 2$

Case II:
$\frac{1}{3} = \frac{2 k + 1}{7 k + 1}$
$\Rightarrow (7 k + 1) = 6 k + 3$
$\Rightarrow k = 2$

Case III:
$\frac{k}{2 (k + 1)} = \frac{2 k + 1}{7 k + 1}$
$\Rightarrow k (7 k + 1) = (2 k + 1) \times 2 (k + 1)$
$\Rightarrow 7 k^{2} + k = (2 k + 1) (2 k + 2)$
$\Rightarrow 7 k^{2} + k = 4 k^{2} + 4 k + 2 k + 2$
$\Rightarrow 3 k^{2} - 5 k - 2 = 0$
$\Rightarrow 3 k^{2} - 6 k + k - 2 = 0$
$\Rightarrow 3 k (k - 2) + 1 (k - 2) = 0$
$\Rightarrow (3 k + 1) (k - 2) = 0$
$\Rightarrow k = 2 or k = \frac{- 1}{3}$
Hence, the given system of equations has an infinite number of solutions when k is equal to 2.

Page No 123:

Question 18:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$5 x + 2 y = 2 k, 2 (k + 1) x + k y = (3 k + 4) .$

Answer:

The given system of equations:
5x + 2y = 2k
⇒ 5x + 2y − 2k= 0 ...(i)
And, 2(k + 1)x + ky = (3k + 4)
⇒ 2(k + 1)x + ky − (3k + 4) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 5, b₁= 2, c₁ = −2k and a₂ = 2(k + 1), b₂= k, c₂ = −(3k + 4)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{5}{2 (k + 1)} = \frac{2}{k} = \frac{- 2 k}{- (3 k + 4)}$
$\Rightarrow \frac{5}{2 (k + 1)} = \frac{2}{k} = \frac{2 k}{(3 k + 4)}$

Now, we have the following three cases:
Case I:
$\frac{5}{2 (k + 1)} = \frac{2}{k}$
$\Rightarrow 2 \times 2 (k + 1) = 5 k \Rightarrow 4 (k + 1) = 5 k$
$\Rightarrow 4 k + 4 = 5 k \Rightarrow k = 4$

Case II:
$\frac{2}{k} = \frac{2 k}{(3 k + 4)}$
$\Rightarrow 2 k^{2} = 2 \times (3 k + 4)$
$\Rightarrow 2 k^{2} = 6 k + 8 \Rightarrow 2 k^{2} - 6 k - 8 = 0$
$\Rightarrow 2 (k^{2} - 3 k - 4) = 0$
$\Rightarrow k^{2} - 4 k + k - 4 = 0$
$\Rightarrow k (k - 4) + 1 (k - 4) = 0$
$\Rightarrow (k + 1) (k - 4) = 0$
$\Rightarrow (k + 1) = 0 or (k - 4) = 0$
$\Rightarrow k = - 1 or k = 4$

Case III:
$\frac{5}{2 (k + 1)} = \frac{2 k}{3 k + 4}$
$\Rightarrow 15 k + 20 = 4 k^{2} + 4 k$
$\Rightarrow 4 k^{2} - 11 k - 20 = 0$
$\Rightarrow 4 k^{2} - 16 k + 5 k - 20 = 0$
$\Rightarrow 4 k (k - 4) + 5 (k - 4) = 0$
$\Rightarrow (k - 4) (4 k + 5) = 0$
$\Rightarrow k = 4 or k = \frac{- 5}{4}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 4.

Page No 123:

Question 19:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$(k - 1) x - y = 5, (k + 1) x + (1 - k) y = (3 k + 1) .$

Answer:

The given system of equations:
(k − 1)x − y = 5
⇒ (k − 1)x − y − 5 = 0 ...(i)
And, (k + 1)x + (1 − k)y = (3k + 1)
⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = (k − 1), b₁= −1, c₁ = −5 and a₂ = (k + 1), b₂= (1 − k), c₂ = −(3k + 1)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$i . e . \frac{(k - 1)}{(k + 1)} = \frac{- 1}{- (k - 1)} = \frac{- 5}{- (3 k + 1)}$
$\Rightarrow \frac{(k - 1)}{(k + 1)} = \frac{1}{(k - 1)} = \frac{5}{(3 k + 1)}$

Now, we have the following three cases:
Case I:
$\frac{(k - 1)}{(k + 1)} = \frac{1}{(k - 1)}$
$\Rightarrow {(k - 1)}^{2} = (k + 1)$
$\Rightarrow k^{2} + 1 - 2 k = k + 1$
$\Rightarrow k^{2} - 3 k = 0 \Rightarrow k (k - 3) = 0 \Rightarrow k = 0 or k = 3$

Case II:
$\frac{1}{(k - 1)} = \frac{5}{(3 k + 1)}$
$\Rightarrow 3 k + 1 = 5 (k - 1)$
$\Rightarrow 3 k + 1 = 5 k - 5$
$\Rightarrow 2 k = 6 \Rightarrow k = 3$

Case III:
$\frac{(k - 1)}{(k + 1)} = \frac{5}{(3 k + 1)}$
$\Rightarrow (3 k + 1) (k - 1) = 5 (k + 1)$
$\Rightarrow 3 k^{2} + k - 3 k - 1 = 5 k + 5$
$\Rightarrow 3 k^{2} - 2 k - 5 k - 1 - 5 = 0$
$\Rightarrow 3 k^{2} - 7 k - 6 = 0$
$\Rightarrow 3 k^{2} - 9 k + 2 k - 6 = 0$
$\Rightarrow 3 k (k - 3) + 2 (k - 3) = 0$
$\Rightarrow (k - 3) (3 k + 2) = 0$
$\Rightarrow (k - 3) = 0 or (3 k + 2) = 0$
$\Rightarrow k = 3 or k = \frac{- 2}{3}$

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.

Page No 123:

Question 20:

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
$(k - 3) x + 3 y = k, k x + k y = 12 .$

Answer:

The given system of equations can be written as
$(k - 3) x + 3 y - k = 0 . . . . . (i) k x + k y - 12 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = k - 3, b_{1} = 3, c_{1} = - k and a_{2} = k, b_{2} = k, c_{2} = - 12$
For the given system of linear equations to have an infinite number of solutions
, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{k - 3}{k} = \frac{3}{k} = \frac{- k}{- 12} \Rightarrow \frac{k - 3}{k} = \frac{3}{k} and \frac{3}{k} = \frac{- k}{- 12} \Rightarrow k - 3 = 3 and k^{2} = 36$
$\Rightarrow k = 6 and k = \pm 6 \Rightarrow k = 6$
Hence, k = 6.

Page No 123:

Question 21:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$(a - 1) x + 3 y = 2, 6 x + (1 - 2 b) y = 6 .$

Answer:

The given system of equations:
(a − 1)x + 3y = 2
⇒ (a − 1)x + 3y − 2 = 0 ...(i)
and 6x + (1 − 2b)y = 6
⇒ 6x + (1 − 2b)y − 6= 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
where, a₁ = (a − 1), b₁= 3, c₁ = −2 and a₂ = 6, b₂= (1 − 2b), c₂ = −6
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{(a - 1)}{6} = \frac{3}{(1 - 2 b)} = \frac{- 2}{- 6}$
$\Rightarrow \frac{a - 1}{6} = \frac{3}{(1 - 2 b)} = \frac{1}{3}$
$\Rightarrow \frac{a - 1}{6} = \frac{1}{3} and \frac{3}{(1 - 2 b)} = \frac{1}{3}$
⇒ 3a − 3 = 6 and 9 = 1 − 2b
⇒ 3a = 9 and 2b = −8
⇒ a = 3 and b = −4
∴ a = 3 and b = −4

Page No 124:

Question 22:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$(2 a - 1) x + 3 y = 5, 3 x + (b - 1) y = 2 .$

Answer:

The given system of equations:
(2a − 1)x + 3y = 5
⇒ (2a − 1)x + 3y − 5 = 0 ....(i)
And, 3x + (b − 1)y = 2
⇒ 3x + (b − 1)y − 2 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = (2a − 1), b₁= 3, c₁ = −5 and a₂ = 3, b₂= (b − 1), c₂ = −2
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{(2 a - 1)}{3} = \frac{3}{(b - 1)} = \frac{- 5}{- 2}$
$\Rightarrow \frac{(2 a - 1)}{3} = \frac{3}{(b - 1)} = \frac{5}{2}$
$\Rightarrow \frac{(2 a - 1)}{3} = \frac{5}{2} and \frac{3}{(b - 1)} = \frac{5}{2}$
⇒ 2(2a − 1) = 15 and 6 = 5(b − 1)
⇒ 4a − 2 = 15 and 6 = 5b − 5
⇒ 4a = 17 and 5b = 11
$\Rightarrow a = \frac{17}{4} and b = \frac{11}{5}$
∴ a = $\frac{17}{4}$ and b = $\frac{11}{5}$

Page No 124:

Question 23:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$2 x - 3 y = 7, (a + b) x - (a + b - 3) y = 4 a + b .$

Answer:

The given system of equations:
2x − 3y = 7
⇒ 2x − 3y − 7 = 0                                ....(i)
And, (a + b)x − (a + b − 3)y = 4a + b
⇒ (a + b)x − (a + b − 3)y − (4a + b) = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= −3, c₁ = −7 and a₂ = (a + b), b₂= −(a + b − 3), c₂ = −(4a + b)
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{2}{a + b} = \frac{- 3}{- (a + b - 3)} = \frac{- 7}{- (4 a + b)}$
$\Rightarrow \frac{2}{a + b} = \frac{3}{(a + b - 3)} = \frac{7}{(4 a + b)}$
$\Rightarrow \frac{2}{a + b} = \frac{7}{(4 a + b)} and \frac{3}{(a + b - 3)} = \frac{7}{(4 a + b)}$
⇒ 2(4a + b) = 7(a + b) and 3(4a + b) = 7(a + b − 3)
⇒ 8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b − 21
⇒ a = 5b                  ....(iii)
And, 5a = 4b − 21     ....(iv)
On substituting a = 5b in (iv), we get:
25b = 4b − 21
⇒ 21b = −21
⇒ b = −1
On substituting b = −1 in (iii), we get:
a = 5(−1) = −5
∴ a = −5 and b = −1

Page No 124:

Question 24:

Find the values of a and b for which each of the following systems of linear equations has an infinite number of solutions:
$2 x + 3 y = 7, (a + b + 1) x + (a + 2 b + 2) y = 4 (a + b) + 1 .$

Answer:

The given system of equations:
2x + 3y = 7
⇒ 2x + 3y − 7 = 0                                   ....(i)
And, (a + b + 1)x + (a + 2b + 2)y = 4(a + b) + 1
⇒ (a + b + 1)x + (a + 2b + 2)y − [4(a + b) + 1] = 0    ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁= 3, c₁ = −7 and a₂ = (a + b + 1), b₂= (a + 2b + 2), c₂ = −[4(a + b) + 1]
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{2}{(a + b + 1)} = \frac{3}{(a + 2 b + 2)} = \frac{- 7}{- [4 (a + b) + 1]}$
$\Rightarrow \frac{2}{(a + b + 1)} = \frac{3}{(a + 2 b + 2)} = \frac{7}{4 (a + b) + 1}$
$\Rightarrow \frac{2}{(a + b + 1)} = \frac{3}{(a + 2 b + 2)} and \frac{3}{(a + 2 b + 2)} = \frac{7}{4 (a + b) + 1}$

⇒ 2(a + 2b + 2) = 3(a + b+ 1) and 3[4(a + b) + 1] = 7(a + 2b + 2)
⇒ 2a + 4b + 4 = 3a + 3b + 3 and 3(4a + 4b + 1) = 7a + 14b + 14
⇒ a − b − 1 = 0 and 12a + 12b + 3 = 7a + 14b + 14
⇒ a − b = 1 and 5a − 2b = 11
a = (b + 1)     ....(iii)
5a − 2b = 11     ....(iv)
On substituting a = (b + 1) in (iv), we get:
5(b + 1) − 2b = 11
⇒ 5b + 5 − 2b = 11
⇒ 3b = 6
⇒ b = 2
On substituting b = 2 in (iii), we get:
a = 3
∴ a = 3 and b = 2

Page No 124:

Question 25:

Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
$2 x + 3 y = 7 (a + b) x + (2 a - b) y = 21$ [CBSE 2001]

Answer:

The given system of equations can be written as
$2 x + 3 y - 7 = 0 . . . . . (i) (a + b) x + (2 a - b) y - 21 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 2, b_{1} = 3, c_{1} = - 7 and a_{2} = a + b, b_{2} = 2 a - b, c_{2} = - 21$
For the given system of linear equations to have an infinite number of solutions, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{a + b} = \frac{3}{2 a - b} = \frac{- 7}{- 21} \Rightarrow \frac{2}{a + b} = \frac{- 7}{- 21} = \frac{1}{3} and \frac{3}{2 a - b} = \frac{- 7}{- 21} = \frac{1}{3} \Rightarrow a + b = 6 and 2 a - b = 9$
Adding $a + b = 6 and 2 a - b = 9$ , we get
$3 a = 15 \Rightarrow a = \frac{15}{3} = 5$
Now, substituting a = 5 in a + b = 6, we have
$5 + b = 6 \Rightarrow b = 6 - 5 = 1$
Hence, a = 5 and b = 1.

Page No 124:

Question 26:

Find the values of a and b for which the following system of linear equations has an infinite number of solutions:
$2 x + 3 y = 7 2 a x + (a + b) y = 28$ [CBSE 2001]

Answer:

The given system of equations can be written as
$2 x + 3 y - 7 = 0 . . . . . (i) 2 a x + (a + b) y - 28 = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = 2, b_{1} = 3, c_{1} = - 7 and a_{2} = 2 a, b_{2} = a + b, c_{2} = - 28$
For the given system of linear equations to have an infinite number of solutions, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{2 a} = \frac{3}{a + b} = \frac{- 7}{- 28} \Rightarrow \frac{2}{2 a} = \frac{- 7}{- 28} = \frac{1}{4} and \frac{3}{a + b} = \frac{- 7}{- 28} = \frac{1}{4} \Rightarrow a = 4 and a + b = 12$
Substituting a = 4 in a + b = 12, we get
$4 + b = 12 \Rightarrow b = 12 - 4 = 8$
Hence, a = 4 and b = 8.

Page No 124:

Question 27:

Find the value of k for which each of the following system of equations has no solution: $8 x + 5 y = 9, k x + 10 y = 15 .$

Answer:

The given system of equations:
8x + 5y = 9
8x + 5y − 9 = 0 ....(i)
kx + 10y = 15
kx + 10y − 15= 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 8, b₁= 5, c₁ = −9 and a₂ = k, b₂= 10, c₂ = −15
In order that the given system has no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$i . e . \frac{8}{k} = \frac{5}{10} \neq \frac{- 9}{- 15} i . e . \frac{8}{k} = \frac{1}{2} \neq \frac{3}{5}$
$\frac{8}{k} = \frac{1}{2}$ and $\frac{8}{k} \neq \frac{3}{5}$
$\Rightarrow k = 16 and k \neq \frac{40}{3}$
Hence, the given system of equations has no solution when k is equal to 16.

Page No 124:

Question 28:

Find the value of k for which each of the following system of equations has no solution:
$k x + 3 y = 3, 12 x + k y = 6 .$

Answer:

The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0 ....(i)
12x + ky = 6
12x + ky − 6= 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 3, c₁ = −3 and a₂ = 12, b₂= k, c₂ = −6
In order that the given system of equations has no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
i.e. $\frac{k}{12} = \frac{3}{k} \neq \frac{- 3}{- 6}$
$\frac{k}{12} = \frac{3}{k}$ and $\frac{3}{k} \neq \frac{1}{2}$
$\Rightarrow k^{2} = 36 and k \neq 6$
$\Rightarrow k = \pm 6 and k \neq 6$
Hence, the given system of equations has no solution when k is equal to −6.

Page No 124:

Question 29:

Find the value of k for which each of the following system of equations has no solution:
$3 x - y - 5 = 0, 6 x - 2 y + k = 0, (k \neq 0) .$

Answer:

The given system of equations:
3x − y − 5 = 0 ...(i)
And, 6x − 2y + k = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁= −1, c₁ = −5 and a₂ = 6, b₂= −2, c₂ = k
In order that the given system of equations has no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
i.e. $\frac{3}{6} = \frac{- 1}{- 2} \neq \frac{- 5}{k}$
⇒ $\frac{- 1}{- 2} \neq \frac{- 5}{k} \Rightarrow k \neq - 10$
Hence, equations (i) and (ii) will have no solution if $k \neq - 10$ .

Page No 124:

Question 30:

Find the value of k for which the following system of linear equations has no solutions:
$k x + 3 y = k - 3 12 x + k y = k$

Answer:

The given system of equations can be written as
$k x + 3 y + 3 - k = 0 . . . . . (i) 12 x + k y - k = 0 . . . . . (i i)$
This system is of the form
$a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0$
where $a_{1} = k, b_{1} = 3, c_{1} = 3 - k and a_{2} = 12, b_{2} = k, c_{2} = - k$
For the given system of linear equations to have no solution, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{k}{12} = \frac{3}{k} \neq \frac{3 - k}{- k} \Rightarrow \frac{k}{12} = \frac{3}{k} and \frac{3}{k} \neq \frac{3 - k}{- k} \Rightarrow k^{2} = 36 and - 3 \neq 3 - k$
$\Rightarrow k = \pm 6 and k \neq 6 \Rightarrow k = - 6$
Hence, $k = - 6$ .

Page No 124:

Question 31:

Find the value of k for which the system of equations
$5 x - 3 y = 0, 2 x + k y = 0$
has a nonzero solution.

Answer:

The given system of equations:
5x − 3y = 0 ....(i)
2x + ky = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 5, b₁= −3, c₁ = 0 and a₂ = 2, b₂= k, c₂ = 0
For a non-zero solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$
$\Rightarrow \frac{5}{2} = \frac{- 3}{k}$
$\Rightarrow 5 k = - 6 \Rightarrow k = \frac{- 6}{5}$
Hence, the required value of k is $\frac{- 6}{5}$ .

Page No 145:

Question 1:

5 chairs and 4 tables together cost ₹ 5,600, while 4 chairs and 3 tables together cost
₹ 4,340. Find the cost of a chair and that of a table.

Answer:

Let the cost of a chair be ₹ x and that of a table be ₹ y. Then
$5 x + 4 y = 5600 . . . . . (i) 4 x + 3 y = 4340 . . . . . (i i)$
Multiplying (i) by 3 and (ii) by 4, we get
$15 x - 16 x = 16800 - 17360 \Rightarrow - x = - 560 \Rightarrow x = 560$
Substituting x = 560 in (i), we have
$5 \times 560 + 4 y = 5600 \Rightarrow 4 y = 5600 - 2800 \Rightarrow y = \frac{2800}{4} = 700$
Hence, the cost of a chair and that of a table are respectively ₹ 560 and ₹ 700.

Page No 145:

Question 2:

23 spoons and 17 forks together cost ₹1,770, while 17 spoons and 23 forks together
cost ₹1,830. Find the cost of a spoon and that of a fork.

Answer:

Let the cost of a spoon be ₹x and that of a fork be ₹y. Then
$23 x + 17 y = 1770 . . . . . (i) 17 x + 23 y = 1830 . . . . . (i i)$
Adding (i) and (ii), we get
$40 x + 40 y = 3600 \Rightarrow x + y = 90 . . . . . (i i i)$
Now, subtracting (ii) from (i), we get
$6 x - 6 y = - 60 \Rightarrow x - y = - 10 . . . . . (i v)$
Adding (iii) and (iv), we get
$2 x = 80 \Rightarrow x = 40$
Substituting x = 40 in (iii), we get
$40 + y = 90 \Rightarrow y = 50$
Hence, the cost of a spoon that of a fork are ₹40 and ₹50 respectively.

Page No 146:

Question 3:

A lady has only 25-paisa and 50-paisa coins in her purse. If she has 50 coins in all
totalling ₹19.50, how many coins of each kind does she have?

Answer:

Let the x and y be the number of 50-paisa and 25-paisa conis respectively. Then
$x + y = 50 . . . . . (i) 0.5 x + 0.25 y = 19.50 . . . . . (i i)$
Multiplying (ii) by 2 and subtracting it from (i), we get
$0.5 y = 50 - 39 \Rightarrow y = \frac{11}{0.5} = 22$
Subtracting y = 22 in (i), we get
$x + 22 = 50 \Rightarrow x = 50 - 22 = 28$
Hence, the number of 25-paisa and 50-paisa conis are 22 and 28 respectively.

Page No 146:

Question 4:

The sum of two numbers is 137 and their difference is 43. Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then, we have:
x + y = 137 ...(i)
x − y = 43 ...(ii)
On adding (i) and (ii), we get:
2x = 180 ⇒ x = 90
On substituting x = 90 in (i), we get:
90 + y = 137
⇒ y = (137 − 90) = 47
Hence, the required numbers are 90 and 47.

Page No 146:

Question 5:

Find two numbers such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

Answer:

Let the first number be x and the second number be y.
Then, we have:
2x + 3y = 92                       ....(i)
4x − 7y = 2                         ....(ii)
On multiplying (i) by 7 and (ii) by 3, we get:
14x + 21y = 644                 ....(iii)
12x − 21y = 6                     ....(iv)
On adding (iii) and (iv), we get:
26x = 650
⇒ x = 25
On substituting x = 25 in (i), we get:
2 × 25 + 3y = 92
⇒ 50 + 3y = 92
⇒ 3y = (92 − 50) = 42
⇒ y = 14
Hence, the first number is 25 and the second number is 14.

Page No 146:

Question 6:

Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.

Answer:

Let the first number be x and the second number be y.
Then, we have:
3x + y = 142 ....(i)
4x − y = 138 ....(ii)
On adding (i) and (ii), we get:
7x = 280
⇒ x = 40
On substituting x = 40 in (i), we get:
3 × 40 + y = 142
⇒ y = (142 − 120) = 22
⇒ y = 22
Hence, the first number is 40 and the second number is 22.

Page No 146:

Question 7:

If 45 is subtracted from twice the greater of two numbers, it result in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.

Answer:

Let the greater number be x and the smaller be y.
Then, we have:
2x − 45 = y or 2x − y = 45             .... (i)
2y − 21 = x or −x + 2y = 21           ....(ii)
On multiplying (i) by 2, we get:
4x − 2y = 90                                      ....(iii)
On adding (ii) and (iii), we get:
3x = (90 + 21) = 111
⇒ x = 37
On substituting x = 37 in (i), we get:
2 × 37 − y = 45
⇒ 74 − y = 45
⇒ y = (74 − 45) = 29
Hence, the greater number is 37 and the smaller number is 29.

Page No 146:

Question 8:

If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.

Answer:

We know:
Dividend = Divisor × Quotient + Remainder

Let the larger number be x and the smaller be y.
Then, we have:
3x = y × 4 + 8 or 3x − 4y = 8 ....(i)
5y = x × 3 + 5 or −3x + 5y = 5 ....(ii)
On adding (i) and (ii), we get:
y = (8 + 5) = 13
On substituting y = 13 in (i) we get:
3x − 4 × 13 = 8
⇒ 3x = (8 + 52) = 60
⇒ x = 20
Hence, the larger number is 20 and the smaller number is 13.

Page No 146:

Question 9:

If 2 is added to each of two given numbers, their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

Answer:

Let the required numbers be x and y.
Now, we have:
$\frac{x + 2}{y + 2} = \frac{1}{2}$
By cross multiplication, we get:
2x + 4 = y + 2
⇒ 2x − y = −2                  ....(i)
Again, we have:
$\frac{x - 4}{y - 4} = \frac{5}{11}$
By cross multiplication, we get:
11x − 44 = 5y − 20
⇒ 11x − 5y = 24                ....(ii)
On multiplying (i) by 5, we get:
10x − 5y = −10                 ....(iii)
On subtracting (iii) from (ii), we get:
x = (24 + 10) = 34
On substituting x = 34 in (i), we get:
2 × 34 − y = −2
⇒ 68 − y = −2
⇒ y = (68 + 2) = 70
Hence, the required numbers are 34 and 70.

Page No 146:

Question 10:

The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 14 or x = 14 + y ....(i)
x² − y² = 448 ....(ii)
On substituting x = 14 + y in (ii), we get:
(14 + y)² − y² = 448
⇒ 196 + y² + 28y − y² = 448
⇒ 196 + 28y = 448
⇒ 28y = (448 − 196) = 252
⇒ $y = \frac{252}{28} = 9$
On substituting y = 9 in (i), we get:
x = 14 + 9 = 23
Hence, the required numbers are 23 and 9.

Page No 146:

Question 11:

The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
x + y = 12 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 18
⇒ 10y + x − 10x − y = 18
⇒ 9y − 9x = 18
⇒ y − x = 2 ....(ii)
On adding (i) and (ii), we get:
2y = 14
⇒ y = 7
On substituting y = 7 in (i), we get:
x + 7 = 12
⇒ x = (12 − 7) = 5
Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.

Page No 146:

Question 12:

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
10x + y = 7(x + y)
10x + y = 7x + 7y or 3x − 6y = 0       ....(i)
Number obtained on reversing its digits = (10y + x)
(10x + y) − 27 = 10y + x
⇒ 10x − x + y − 10y = 27
⇒ 9x − 9y = 27
⇒ 9(x − y) = 27
⇒ x − y = 3                        ....(ii)
On multiplying (ii) by 6, we get:
6x − 6y = 18                      ....(iii)
On subtracting (i) from (iii), we get:
3x = 18
⇒ x = 6
On substituting x = 6 in (i), we get:
3 × 6 − 6y = 0
⇒ 18 − 6y = 0
⇒ 6y = 18
⇒ y = 3
Number = (10x + y) = 10 × 6 + 3 = 60 + 3 = 63
Hence, the required number is 63.

Page No 146:

Question 13:

The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) − (10x + y) = 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16
⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15
⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.

Page No 146:

Question 14:

A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
∴ 10x + y = 4(x + y) + 3
⇒10x + y = 4x + 4y + 3
⇒ 6x − 3y = 3
⇒ 2x − y = 1 ....(i)

Again, we have:
10x + y + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ x − y = −2 ....(ii)
On subtracting (ii) from (i), we get:
x = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = 1
⇒ y = 6 − 1 = 5
Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35
Hence, the required number is 35.

Page No 146:

Question 15:

A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

Answer:

We know:
Dividend = (Divisor × Quotient) + Remainder

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
∴ 10x + y = (x + y) × 6 + 0
⇒ 10x − 6x + y − 6y = 0
⇒ 4x − 5y = 0                     ....(i)
Number obtained on reversing its digits = (10y + x)
∴ 10x + y − 9 = 10y + x
⇒ 9x − 9y = 9
⇒ 9(x − y) = 9
⇒ x − y = 1                        ....(ii)
On multiplying (ii) by 5, we get:
5x − 5y = 5                        ....(iii)
On subtracting (i) from (iii), we get:
x = 5
On substituting x = 5 in (i), we get:
4 × 5 − 5y = 0
⇒ 20 − 5y = 0
⇒ y = 4
∴ Number = (10x + y) = 10 × 5 + 4 = 50 + 4 = 54
Hence, the required number is 54.

Page No 147:

Question 16:

A two-digit number is such that the product of its digit is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Then, we have:
xy = 35                      ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) + 18 = 10y + x
⇒ 9x − 9y = −18
⇒ 9(y − x) = 18
⇒ y − x = 2                 ....(ii)

We know:
(y + x)² − (y − x)² = 4xy
⇒ $(y + x) = \pm \sqrt{{(y - x)}^{2} + 4 x y}$
⇒ $(y + x) = \pm \sqrt{4 + 4 \times 35} = \pm \sqrt{144} = \pm 12$
∴ y + x = 12              .....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2y = 2 + 12 = 14
⇒ y = 7
On substituting y = 7 in (ii), we get:
7 − x = 2
⇒ x = (7 − 2) = 5
∴ Number = (10x + y) = 10 × 5 + 7 = 50 + 7 = 57
Hence, the required number is 57.

Page No 147:

Question 17:

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Answer:

Let the tens and the units digits of the required number be x and y, respectively.
Then, we have:
xy = 18                       ....(i)
Required number = (10x + y)
Number obtained on reversing its digits = (10y + x)
∴ (10x + y) − 63 = 10y + x
⇒ 9x − 9y = 63
⇒ 9(x − y) = 63
⇒ x − y = 7        ....(ii)

We know:
(x + y)² − (x − y)² = 4xy
⇒ $(x + y) = \pm \sqrt{{(x - y)}^{2} + 4 x y}$
$\Rightarrow (x + y) = \pm \sqrt{49 + 4 \times 18} = \pm \sqrt{49 + 72} = \pm \sqrt{121} = \pm 11$
∴ x + y = 11              ....(iii) (∵ x and y cannot be negative)
On adding (ii) and (iii), we get:
2x = 7 + 11 = 18
⇒ x = 9
On substituting x = 9 in (ii), we get:
9 − y = 7
⇒ y = (9 − 7) = 2
∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92
Hence, the required number is 92.

Page No 147:

Question 18:

The sum of a two-digit number and the number obtained by reversing the order of its digits
is 121, and the two digits differ by 3. Find the number.

Answer:

Let x be the ones digit and y be the tens digit. Then
Two digit number before reversing = 10y + x
Two digit number after reversing = 10x + y
As per the question
$(10 y + x) + (10 x + y) = 121 \Rightarrow 11 x + 11 y = 121 \Rightarrow x + y = 11 . . . . . (i)$
Since the digits differ by 3, so
$x - y = 3 . . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 14 \Rightarrow x = 7$
Putting x = 7 in (i), we get
$7 + y = 11 \Rightarrow y = 4$
Changing the role of x and y, x = 4 and y = 7
Hence, the two-digit number is 74 or 47.

Page No 147:

Question 19:

The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes $\frac{3}{4}$ . Find the fraction.

Answer:

Let the required fraction be $\frac{x}{y}$ .
Then, we have:
x + y = 8                        ....(i)
And, $\frac{x + 3}{y + 3} = \frac{3}{4}$
⇒ 4(x + 3) = 3(y + 3)
⇒ 4x + 12 = 3y + 9
⇒ 4x − 3y = −3              ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 24                  ....(iii)
On adding (ii) and (iii), we get:
7x = 21
⇒ x = 3
On substituting x = 3 in (i), we get:
3 + y = 8
⇒ y = (8 − 3) = 5
∴ x = 3 and y = 5
Hence, the required fraction is $\frac{3}{5}$ .

Page No 147:

Question 20:

If 2 is added to the numerator of a fraction, it reduces to $(\frac{1}{2})$ and if 1 is subtracted from the denominator, it reduces to $(\frac{1}{3})$ . Find the fraction.

Answer:

Let the required fraction be $\frac{x}{y}$ .
Then, we have:
$\frac{x + 2}{y} = \frac{1}{2}$
⇒ 2(x + 2) = y
⇒ 2x + 4 = y
⇒ 2x − y = −4 .....(i)

Again, $\frac{x}{y - 1} = \frac{1}{3}$
⇒ 3x = 1(y − 1)
⇒ 3x − y = −1 .....(ii)
On subtracting (i) from (ii), we get:
x = (−1 + 4) = 3
On substituting x = 3 in (i), we get:
2 × 3 − y = −4
⇒ 6 − y = −4
⇒ y = (6 + 4) = 10
∴ x = 3 and y = 10
Hence, the required fraction is $\frac{3}{10}$ .

Page No 147:

Question 21:

The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes $\frac{3}{4}$ . Find the fraction.

Answer:

Let the required fraction be $\frac{x}{y}$ .
Then, we have:
y = x + 11
⇒ y − x = 11                      ....(i)
Again, $\frac{x + 8}{y + 8} = \frac{3}{4}$
⇒ 4(x + 8) =3(y + 8)
⇒ 4x + 32 = 3y + 24
⇒ 4x − 3y = −8                    ....(ii)
On multiplying (i) by 4, we get:
4y − 4x = 44                        ....(iii)
On adding (ii) and (iii), we get:
y = (−8 + 44) = 36
On substituting y = 36 in (i), we get:
36 − x = 11
⇒ x = (36 − 11) = 25
∴ x = 25 and y = 36
Hence, the required fraction is $\frac{25}{36}$ .

Page No 147:

Question 22:

Find a fraction which becomes $(\frac{1}{2})$ when 1 is subtracted from the numerator and 2 is added to the denominator, and the fraction becomes $(\frac{1}{3})$ when 7 is subtracted from the numerator and 2 is subtracted from the denominator.

Answer:

Let the required fraction be $\frac{x}{y}$ .
Then, we have:
$\frac{x - 1}{y + 2} = \frac{1}{2}$
⇒ 2(x − 1) = 1(y + 2)
⇒ 2x − 2 = y + 2
⇒ 2x − y = 4 ....(i)

Again, $\frac{x - 7}{y - 2} = \frac{1}{3}$
⇒ 3(x − 7) = 1(y − 2)
⇒ 3x − 21 = y − 2
⇒ 3x − y = 19 ....(ii)
On subtracting (i) from (ii), we get:
x = (19 − 4) = 15
On substituting x = 15 in (i), we get:
2 × 15 − y = 4
⇒ 30 − y = 4
⇒ y = 26
∴ x = 15 and y = 26
Hence, the given fraction is $\frac{15}{26}$ .

Page No 147:

Question 23:

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction. [CBSE 2010]

Answer:

Let the fraction be $\frac{x}{y}$ .
As per the question
$x + y = 4 + 2 x \Rightarrow y - x = 4 . . . . . (i)$
After changing the numerator and denominator
New numerator = x + 3
New denominator = y + 3
Therefore
$\frac{x + 3}{y + 3} = \frac{2}{3} \Rightarrow 3 (x + 3) = 2 (y + 3) \Rightarrow 3 x + 9 = 2 y + 6 \Rightarrow 2 y - 3 x = 3 . . . . . (i i)$
Multiplying (i) by 3 and subtracting (ii), we get
$3 y - 2 y = 12 - 3 \Rightarrow y = 9$
Now, putting y = 9 in (i), we get
$9 - x = 4 \Rightarrow x = 9 - 4 = 5$
Hence, the fraction is $\frac{5}{9}$ .

Page No 147:

Question 24:

The sum of two numbers is 16 and the sum of their reciprocals is $\frac{1}{3}$ . Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then, we have:
(x + y) = 16                      ....(i)
And, $\frac{1}{x} + \frac{1}{y} = \frac{1}{3}$               ....(ii)
⇒ $\frac{x + y}{x y} = \frac{1}{3}$
⇒ 3(x + y) = xy
⇒ 3 × 16 = xy      [Since from (i), we have: x + y = 16]
∴ xy = 48                     ....(iii)
We know:
(x − y)² = (x + y)² − 4xy
(x − y)² = (16)² − 4 × 48 = 256 − 192 = 64
∴ (x − y) = $\pm \sqrt{64} = \pm 8$
Since x is larger and y is smaller, we have:
x − y = 8           .....(iv)
On adding (i) and (iv), we get:
2x = 24
⇒ x = 12
On substituting x = 12 in (i), we get:
12 + y = 16 ⇒ y = (16 − 12) = 4
Hence, the required numbers are 12 and 4.

Page No 147:

Question 25:

There are two classrooms A and B. If 10 students are sent from A to B, the number of students in each room becomes the same. If 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in each room.

Answer:

Let the number of students in classroom A be x
Let the number of students in classroom B be y.
If 10 students are transferred from A to B, then we have:
x − 10 = y + 10
⇒ x − y = 20 ...(i)
If 20 students are transferred from B to A, then we have:
2(y − 20) = x + 20
⇒ 2y − 40 = x + 20
⇒ −x + 2y = 60 ...(ii)
On adding (i) and (ii), we get:
y = (20 + 60) = 80
On substituting y = 80 in (i), we get:
x − 80 = 20
⇒ x = (20 + 80) = 100
Hence, the number of students in classroom A is 100 and the number of students in classroom B is 80.

Page No 147:

Question 26:

Taxi charges in a city consist of fixed charges and the remaining depending upon the distance
travelled in kilometres. If a man travels 80 km, he pays ₹1,330, and travelling 90 km, he pays
₹1,490. Find the fixed charges and rate per km.

Answer:

Let fixed charges be ₹x and rate per km be ₹y.
Then as per the question
$x + 80 y = 1330 . . . . . (i) x + 90 y = 1490 . . . . . (i i)$
Subtracting (i) from (ii), we get
$10 y = 160 \Rightarrow y = \frac{160}{10} = 16$
Now, putting y = 16, we have
$x + 80 \times 16 = 1330 \Rightarrow x = 1330 - 1280 = 50$
Hence, the fixed charges be ₹50 and the rate per km is ₹16.

Page No 148:

Question 27:

A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay ₹4,500, whereas a student B who takes food for 30 days, he has to pay ₹5,200. Find the fixed charges per month and the cost of the food per day.

Answer:

Let the fixed charges be ₹x and the cost of food per day be ₹y.
Then as per the question
$x + 25 y = 4500 . . . . . (i) x + 30 y = 5200 . . . . . (i i)$
Subtracting (i) from (ii), we get
$5 y = 700 \Rightarrow y = \frac{700}{5} = 140$
Now, putting y = 140, we have
$x + 25 \times 140 = 4500 \Rightarrow x = 4500 - 3500 = 1000$
Hence, the fixed charges is ₹1000 and the cost of the food per day is ₹140.

Page No 148:

Question 28:

A man invested an amount at 10% per annum and another amount at 8% per annum simple interest. Thus, he received ₹1,350 as annual interest. Had he interchanged the amounts invested, he would have received ₹45 less as interest. What amounts did he invest at different rates?

Answer:

Let the the amounts invested at 10% and 8% be ₹x and ₹y respectively.
Then as per the question
$\frac{x \times 10 \times 1}{100} + \frac{y \times 8 \times 1}{100} = 1350 10 x + 8 y = 135000 . . . . . (i)$
After the amounts interchanged but the rate being the same, we have
$\frac{x \times 8 \times 1}{100} + \frac{y \times 10 \times 1}{100} = 1350 - 45 8 x + 10 y = 130500 . . . . . (i i)$
Adding (i) and (ii) and dividing by 9, we get
$2 x + 2 y = 29500 . . . . . (i i i)$
Subtracting (ii) from (i), we get
$2 x - 2 y = 4500 . . . . . (i v)$
Now, adding (iii) and (iv), we have
$4 x = 34000 \Rightarrow x = \frac{34000}{4} = 8500$
Putting x = 8500 in (iii), we get
$2 \times 8500 + 2 y = 29500 \Rightarrow 2 y = 29500 - 17000 = 12500 \Rightarrow y = \frac{12500}{2} = 6250$
Hence, the amounts invested are ₹8,500 at 10% and ₹6,250 at 8%.

Page No 148:

Question 29:

The monthly incomes of A and B are in the ratio 5 : 4 and their monthly expenditures are in the ratio 7 : 5. If each saves ₹9,000 per month, find the monthly income of each.

Answer:

Let the monthly income of A and B are ₹x and ₹y respectively.
Then as per the question
$\frac{x}{y} = \frac{5}{4} \Rightarrow y = \frac{4 x}{5} . . . . . (i)$
Since each save ₹9,000, so
Expenditure of A = ₹ $(x - 9000)$
Expenditure of B = ₹ $(y - 9000)$
The ratio of expenditures of A and B are in the ratio 7 : 5.
$∴ \frac{x - 9000}{y - 9000} = \frac{7}{5} \Rightarrow 7 y - 63000 = 5 x - 45000 \Rightarrow 7 y - 5 x = 18000 . . . . . (i i)$
From (i), substitute $y = \frac{4 x}{5}$ in (ii) to get
$7 \times \frac{4 x}{5} - 5 x = 18000 \Rightarrow 28 x - 25 x = 90000 \Rightarrow 3 x = 90000 \Rightarrow x = 30000$
Now, putting x = 30000, we get
$y = \frac{4 \times 30000}{5} = 4 \times 6000 = 24000$
Hence, the monthly incomes of A and B are ₹30,000 and ₹24,000.

Page No 148:

Question 30:

A man sold a chair and a table together for ₹1,520, thereby making a profit of 25% on chair and 10% on table. By selling them together for ₹1,535, he would would have made a profit of 10% on the chair and 25% on the table. Find the cost of each.

Answer:

Let the cost price of the chair and table be ₹x and ₹y respectively.
Then as per the question
Selling price of chair + Selling price of table = 1520
$\frac{100 + 25}{100} \times x + \frac{100 + 10}{100} \times y = 1520 \Rightarrow \frac{125}{100} x + \frac{110}{100} y = 1520 \Rightarrow 25 x + 22 y - 30400 = 0 . . . . . (i)$
When the profit on chair and table are 10% and 25% respectively, then

$\frac{100 + 10}{100} \times x + \frac{100 + 25}{100} \times y = 1535 \Rightarrow \frac{110}{100} x + \frac{125}{100} y = 1535 \Rightarrow 22 x + 25 y - 30700 = 0 . . . . . (i i)$
Solving (i) and (ii) by cross multiplication, we get
$\frac{x}{(22) (- 30700) - (25) (- 30400)} = \frac{y}{(- 30400) (22) - (- 30700) (25)} = \frac{1}{(25) (25) - (22) (22)} \Rightarrow \frac{x}{7600 - 6754} = \frac{y}{7675 - 6688} = \frac{100}{3 \times 47} \Rightarrow \frac{x}{846} = \frac{y}{987} = \frac{100}{3 \times 47} \Rightarrow x = \frac{100 \times 846}{3 \times 47}, y = \frac{100 \times 987}{3 \times 47}$
$\Rightarrow x = 600, y = 700$
Hence, the cost of chair and table are ₹600 and ₹700 respectively.

Page No 148:

Question 31:

Points A and B are 70 km apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7 hours. But, if they travel towards each other, they meet in 1 hour. Find the speed of each car.

Answer:

Let X and Y be the cars starting from points A and B, respectively and let their speeds be x km/h and y km/h, respectively.
Then, we have the following cases:
Case I: When the two cars move in the same direction
In this case, let the two cars meet at point M.

Distance covered by car X in 7 hours = 7x km
Distance covered by car Y in 7 hours = 7y km
∴ AM = (7x) km and BM = (7y) km
⇒ (AM − BM) = AB
⇒ (7x − 7y) = 70
⇒ 7(x − y) = 70
⇒ (x − y) = 10 ....(i)

Case II: When the two cars move in opposite directions
In this case, let the two cars meet at point N.

Distance covered by car X in 1 hour = x km
Distance covered by car Y in 1 hour = y km
∴ AN = x km and BN = y km
⇒ AN + BN = AB
⇒ x + y = 70 ....(ii)
On adding (i) and (ii), we get:
2x = 80
⇒ x = 40
On substituting x = 40 in (i), we get:
40 − y = 10
⇒ y = (40 − 10) = 30
Hence, the speed of car X is 40 km/h and the speed of car Y is 30 km/h.

Page No 148:

Question 32:

A train covered a certain distance at a uniform speed. If the train had been 5 kmph faster, it would have taken 3 hours less than the scheduled time. And, If the train were slower by 4 kmph, it would have taken 3 hours more than the scheduled time. Find the length of the journey.

Answer:

Let the original speed be x kmph and let the time taken to complete the journey be y hours.
∴ Length of the whole journey = (xy) km
Case I:
When the speed is (x + 5) kmph and the time taken is (y − 3) hrs:
Total journey = (x + 5) (y − 3) km
⇒ (x + 5) (y − 3) = xy
⇒ xy + 5y − 3x − 15 = xy
⇒ 5y − 3x = 15 ...(i)
Case II:
When the speed is (x − 4) kmph and the time taken is (y + 3) hrs:
Total journey = (x − 4) (y + 3) km
⇒ (x − 4) (y + 3) = xy
⇒ xy − 4y + 3x − 12 = xy
⇒ 3x − 4y = 12 ...(ii)
On adding (i) and (ii), we get:
y = 27
On substituting y = 27 in (i), we get:
5 × 27 − 3x = 15
⇒ 135 − 3x = 15
⇒ 3x = 120
⇒ x = 40

∴ Length of the journey = (xy) km = (40 × 27) km = 1080 km

Page No 148:

Question 33:

Abdul travelled 300 km by train and 200 km by taxi taking 5 hours 30 minutes. But, if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.

Answer:

Let the speed of the train and taxi be x km/h and y km/h respectively.
Then as per the question
$\frac{3}{x} + \frac{2}{y} = \frac{11}{200} . . . . . (i)$
When the speeds of the train and taxi are 260 km and 240 km respectively, then
$\frac{260}{x} + \frac{240}{y} = \frac{11}{2} + \frac{6}{60} \Rightarrow \frac{13}{x} + \frac{12}{y} = \frac{28}{100} . . . . . (i i)$
Multiplying (i) by 6 and subtracting (ii) from it, we get
$\frac{18}{x} - \frac{13}{x} = \frac{66}{200} - \frac{28}{100} \Rightarrow \frac{5}{x} = \frac{10}{200} \Rightarrow x = 100$
Putting x = 100 in (i), we have
$\frac{3}{100} + \frac{2}{y} = \frac{11}{200} \Rightarrow \frac{2}{y} = \frac{11}{200} - \frac{3}{100} = \frac{1}{40} \Rightarrow y = 80$
Hence, the speed of the train and that of the taxi are 100 km/h and 80 km/h respectively.

Page No 148:

Question 34:

Places A and B are 160 km apart on a highway. One car starts from A and another car from B at the same time. If they travel in the same direction , they meet in 8 hours. But, if they travel towards each other, they meet in 2 hours. Find the speed of each car.

Answer:

Let the speed of the car A and B be x km/h and y km/h respectively . Let x > y.
Case-1: When they travel in the same direction

From the figure
$A C - B C = 160 \Rightarrow x \times 8 - y \times 8 = 160 \Rightarrow x - y = 20 . . . . . (i)$
Case-2: When they travel in opposite direction

From the figure
$A C + B C = 160 \Rightarrow x \times 2 + y \times 2 = 160 \Rightarrow x + y = 80 . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 100 \Rightarrow x = 50 km / h$
Putting x = 50 in (ii), we have
$50 + y = 80 \Rightarrow y = 80 - 50 = 30 km / h$
Hence, the speeds of the cars are 50 km/h and 30 km/h.

Page No 148:

Question 35:

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Find the speed of the sailor in still water and the speed of the current.

Answer:

Let the speed of the sailor in still water be x km/h and that of the current y km/h.
Speed downstream = (x + y) km/h
Speed upstream = (x − y) km/h
As per the question
$(x + y) \times \frac{40}{60} = 8 \Rightarrow x + y = 12 . . . . . (i)$
When the sailor goes upstream, then
$(x - y) \times 1 = 8 x - y = 8 . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 20 \Rightarrow x = 10$
Putting x = 10 in (i), we have
$10 + y = 12 \Rightarrow y = 2$
Hence, the speeds of the sailor in staill water and the current are 10 km/h and 2 km/h respectively.

Page No 149:

Question 36:

A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Answer:

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Then, we have:
Speed upstream = (x − y)km/hr
Speed downstream = (x + y) km/hr
Time taken to cover 12 km upstream = $\frac{12}{(x - y)}$ hrs
Time taken to cover 40 km downstream = $\frac{40}{(x + y)}$ hrs
Total time taken = 8 hrs
∴ $\frac{12}{(x - y)}$ + $\frac{40}{(x + y)}$ = 8        ....(i)

Again, we have:
Time taken to cover 16 km upstream = $\frac{16}{(x - y)}$ hrs
Time taken to cover 32 km downstream = $\frac{32}{(x + y)}$ hrs
Total time taken = 8 hrs

∴ $\frac{16}{(x - y)}$ + $\frac{32}{(x + y)}$ = 8        ....(ii)
Putting $\frac{1}{(x - y)} = u$ and $\frac{1}{(x + y)} = v$ in (i) and (ii), we get:
12u + 40v = 8
3u + 10v = 2            ....(a)
And, 16u + 32v = 8
⇒ 2u + 4v = 1          ....(b)
On multiplying (a) by 4 and (b) by 10, we get:
12u + 40v = 8                    ....(iii)
And, 20u + 40v = 10            ....(iv)
On subtracting (iii) from (iv), we get:
8u = 2
⇒ $u = \frac{2}{8} = \frac{1}{4}$
On substituting $u = \frac{1}{4}$ in (iii), we get:
40v = 5
⇒ $v = \frac{5}{40} = \frac{1}{8}$
Now, we have:
$u = \frac{1}{4}$
⇒ $\frac{1}{x - y} = \frac{1}{4} \Rightarrow x - y = 4$          ....(v)
$v = \frac{1}{8}$
⇒ $\frac{1}{x + y} = \frac{1}{8} \Rightarrow x + y = 8$             ....(vi)
On adding (v) and (vi), we get:
2x = 12
⇒ x = 6
On substituting x = 6 in (v), we get:
6 − y = 4
⇒ y = (6 − 4) = 2
∴ Speed of the boat in still water = 6 km/h
And, speed of the stream = 2 km/h

Page No 149:

Question 37:

2 men and 5 boys can finish a piece of work in 4 days, while 3 men and 6 boys can finish it in 3 days. Find the time taken by one man alone to finish the work and that taken by one boy alone to finish the work.

Answer:

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days.
∴ One man's one day's work = $\frac{1}{x}$
And, one boy's one day's work = $\frac{1}{y}$
2 men and 5 boys can finish the work in 4 days.
∴ (2 men's one day's work) + (5 boys' one day's work) = $\frac{1}{4}$
⇒ $\frac{2}{x} + \frac{5}{y} = \frac{1}{4}$
⇒ $2 u + 5 v = \frac{1}{4}$               ...(i)           Here, $\frac{1}{x} = u and \frac{1}{y} = v$
Again, 3 men and 6 boys can finish the work in 3 days.
∴ (3 men's one day's work) + (6 boys' one day's work) = $\frac{1}{3}$
⇒ $\frac{3}{x} + \frac{6}{y} = \frac{1}{3}$
⇒ $3 u + 6 v = \frac{1}{3}$              ....(ii)           Here, $\frac{1}{x} = u and \frac{1}{y} = v$
On multiplying (i) by 6 and (ii) by 5, we get:
$12 u + 30 v = \frac{6}{4}$              ....(iii)
$15 u + 30 v = \frac{5}{3}$              ....(iv)
On subtracting (iii) from (iv), we get:
$3 u = (\frac{5}{3} - \frac{6}{4}) = \frac{2}{12} = \frac{1}{6}$
⇒ $u = \frac{1}{6 \times 3} = \frac{1}{18} \Rightarrow \frac{1}{x} = \frac{1}{18} \Rightarrow x = 18$
On substituting $u = \frac{1}{18}$ in (i), we get:
$2 \times \frac{1}{18} + 5 v = \frac{1}{4} \Rightarrow 5 v = (\frac{1}{4} - \frac{1}{9}) = \frac{5}{36}$
⇒ $v = (\frac{5}{36} \times \frac{1}{5}) = \frac{1}{36} \Rightarrow \frac{1}{y} = \frac{1}{36} \Rightarrow y = 36$
Hence, one man alone can finish the work in 18 days and one boy alone can finish the work in 36 days.

Page No 149:

Question 38:

The length of a room exceeds its breadth by 3 metres. If the length in increased by 3 metres and the breadth is decreased by 2 metres, the area remains the same. Find the length and the breadth of the room.

Answer:

Let the length of the room be x metres and the breadth of the room be y metres.
Then, we have:
Area of the room = xy
According to the question, we have:
x = y + 3
⇒ x − y = 3           ....(i)
And, (x + 3) (y − 2) = xy
⇒ xy − 2x + 3y − 6 = xy
⇒ 3y − 2x = 6           ....(ii)
On multiplying (i) by 2, we get:
2x − 2y = 6           ....(iii)
On adding (ii) and (iii), we get:
y = (6 + 6) = 12
On substituting y = 12 in (i), we get:
x − 12 = 3
⇒ x = (3 + 12) =15
Hence, the length of the room is 15 metres and its breadth is 12 metres.

Page No 149:

Question 39:

The area of a rectangle gets reduced by 8 m², when its length is reduced by 5 m and its breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m². Find the length and the breadth of the rectangle.

Answer:

Let the length and the breadth of the rectangle be x m and y m, respectively.
∴ Area of the rectangle = (xy) sq. m

Case 1: When the length is reduced by 5 m and the breadth is increased by 3 m:
New length = (x − 5) m
New breadth = (y + 3) m
∴ New area = (x − 5) (y + 3) sq. m
∴ xy − (x − 5) (y + 3) = 8
⇒ xy − [xy − 5y + 3x − 15] = 8
⇒ xy − xy + 5y − 3x + 15 = 8
⇒ 3x − 5y = 7                .....(i)

Case 2: When the length is increased by 3 m and the breadth is increased by 2 m:
New length = (x + 3) m
New breadth = (y + 2) m
∴ New area = (x + 3) (y + 2) sq. m
∴ (x + 3) (y + 2) − xy = 74
⇒ [xy + 3y + 2x + 6] − xy = 74
⇒ 2x + 3y = 68             .....(ii)
On multiplying (i) by 3 and (ii) by 5, we get:
9x − 15y = 21              .....(iii)
10x + 15y = 340          .....(iv)
On adding (iii) and (iv), we get:
19x = 361
⇒ x = 19
On substituting x = 19 in (iii), we get:
9 × 19 − 15y = 21
⇒ 171 −15y = 21
⇒ 15y = (171 − 21) = 150
⇒ y = 10
Hence, the length is 19 m and the breadth is 10 m.

Page No 149:

Question 40:

The area of a rectangle gets reduced by 67 square metres, when its length is increased by 3 m and breadth is decreased by 4 m.
If the length is reduced by 1 m and breadth is increased by 4 m, the area is increased by 89 square metres. Find the dimensions of the rectangle.

Answer:

Let the length and breadth of the rectangle be x m and y m respectively.
Case1: When length is increased by 3 m and breadth is decreased by 4 m
$x y - (x + 3) (y - 4) = 67 \Rightarrow x y - x y + 4 x - 3 y + 12 = 67 \Rightarrow 4 x - 3 y = 55 . . . . . (i)$
Case2: When length is reduced by 1 m and breadth is increased by 4 m
$(x - 1) (y + 4) - x y = 89 \Rightarrow x y + 4 x - y - 4 - x y = 89 \Rightarrow 4 x - y = 93 . . . . . (i i)$
Subtracting (i) from (ii), we get
$2 y = 38 \Rightarrow y = 19$
Now, putting y = 19 in (ii), we have
$4 x - 19 = 93 \Rightarrow 4 x = 93 + 19 = 112 \Rightarrow x = 28$
Hence, length = 28 m and breadth = 19 m.

Page No 149:

Question 41:

A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket.
One reserved first class ticket from Mumbai to Delhi costs ₹4,150 while one full and one half reserved first class
tickets cost ₹6,255. What is the basic first class full fare and what is the reservation charge?

Answer:

Let the the basic first class full fare be ₹x and the reservation charge be ₹y.
Case 1: One reservation first class full ticket cost ₹4,150
$x + y = 4150 . . . . . (i)$
Case 2: One full and one half reserved first class tickets cost ₹6,255
$(x + y) + (\frac{1}{2} x + y) = 6255 \Rightarrow 3 x + 4 y = 12510 . . . . . (i i)$
Substituting $y = 4150 - x$ from (i) in (ii), we get
$3 x + 4 (4150 - x) = 12510 \Rightarrow 3 x - 4 x + 16600 = 12510 \Rightarrow x = 16600 - 12510 = 4090$
Now, putting x = 4090 in (i), we have
$4090 + y = 4150 \Rightarrow y = 4150 - 4090 = 60$
Hence, cost of basic first class full fare = ₹4,090 and reservation charge = ₹60.

Page No 149:

Question 42:

Five years hence, a man's age will be three times the age of his son. Five years ago, the man was seven times as old as his son.
Find their present ages.

Answer:

Let the the present age of the man be x years and that of his son be y years.
After 5 years man's age = x + 5
After 5 years ago son's age = y + 5
As per the question
$x + 5 = 3 (y + 5) \Rightarrow x - 3 y = 10 . . . . . (i)$
5 years ago man's age = x − 5
5 years ago son's age = y − 5
As per the question
$x - 5 = 7 (y - 5) \Rightarrow x - 7 y = - 30 . . . . . (i i)$
Subtracting (ii) from (i), we have
$4 y = 40 \Rightarrow y = 10$
Putting y = 10 in (i), we get
$x - 3 \times 10 = 10 \Rightarrow x = 10 + 30 = 40$
Hence, man's present age = 40 years and son's present age = 10 years.

Page No 149:

Question 43:

Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the ago of the son. Find the present ages of the man and his son.

Answer:

Let the man's present age be x years.
Let his son's present age be y years.
According to question, we have:
Two years ago:
Age of the man = Five times the age of the son
⇒ (x − 2) = 5(y − 2)
⇒ x − 2 = 5y − 10
⇒ x − 5y = −8 .....(i)
Two years later:
Age of the man = Three times the age of the son + 8
⇒ (x + 2) = 3(y + 2) + 8
⇒ x + 2 = 3y + 6 + 8
⇒ x − 3y = 12 .....(ii)
Subtracting (i) from (ii), we get:
2y = 20
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 5 × 10 = −8
⇒ x − 50 = −8
⇒ x = (−8 + 50) = 42
Hence, the present age of the man is 42 years and the present age of the son is 10 years.

Page No 149:

Question 44:

If twice the son's age in years is added to the father's age, the sum is 70. But, if twice the father's age is added to the son's age, the sum is 95. Find the ages of father and son.

Answer:

Let the father's present age be x years.
Let her son's present age be y years.
Then, we have:
x + 2y = 70                    ....(i)
And, 2x + y = 95               ....(ii)
On multiplying (ii) by 2, we get:
4x + 2y = 190                 ....(iii)
On subtracting (i) from (iii), we get:
3x = 120
⇒ x = 40
On substituting x = 40 in (i), we get:
40 + 2y = 70
⇒ 2y = (70 − 40) = 30
⇒ y = 15
Hence, the father's present age is 40 years and her son's present age is 15 years.

Page No 149:

Question 45:

The present age of a woman is 3 years more than three times the ages of her daughter. Three years hence, the woman's age will be 10 years more than twice the age of her daughter. Find their present ages.

Answer:

Let the woman's present age be x years.
Let her daughter's present age be y years.
Then, we have:
x = 3y + 3
⇒ x − 3y = 3 ....(i)
After three years, we have:
(x + 3) = 2(y + 3) + 10
⇒ x + 3 = 2y + 6 + 10
⇒ x − 2y = 13 ....(ii)
Subtracting (ii) from (i), we get:
−y = (3 − 13) = −10
⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 3
⇒ x − 30 = 3
⇒ x = (3 + 30) = 33
Hence, the woman's present age is 33 years and her daughter's present age is 10 years.

Page No 149:

Question 46:

On selling a tea at 5% loss and a lemon set at 15% gain, a crockery seller gains ₹7. If he sells the tea set at 5% gain and the
lemon set at 10% gain, he gains ₹13. Find the actual price of each of the tea set and the lemon set.

Answer:

Let the actual price of the tea and lemon set be ₹x and ₹y respectively.
When gain is ₹7, then
$\frac{y}{100} \times 15 - \frac{x}{100} \times 5 = 7 \Rightarrow 3 y - x = 140 . . . . . (i)$
When gain is ₹14, then
$\frac{y}{100} \times 5 + \frac{x}{100} \times 10 = 14 \Rightarrow y + 2 x = 280 . . . . . (i i)$
Multiplying (i) by 2 and adding with (ii), we have
$7 y = 280 + 280 \Rightarrow y = \frac{560}{7} = 80$
Putting y = 80 in (ii), we get
$80 + 2 x = 280 \Rightarrow x = \frac{200}{2} = 100$
Hence, actual price of the tea set and lemon set are ₹100 and ₹80 respectively.

Page No 150:

Question 47:

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter.
Mona paid ₹27 for a book kept for 7 days, while Tanvy paid ₹21 for the book she kept for 5 days.
Find the fixed charge and the charge for each extra day.

Answer:

Let the fixed charge be ₹x and the charge for each extra day be ₹y.
In case of Mona, as per the question
$x + 4 y = 27 . . . . . (i)$
In case of Tanvy, as per the question
$x + 2 y = 21 . . . . . (i i)$
Subtracting (ii) from (i), we get
$2 y = 6 \Rightarrow y = 3$
Now, putting y = 3 in (ii), we have
$x + 2 \times 3 = 21 \Rightarrow x = 21 - 6 = 15$
Hence, the fixed charge be ₹15 and the charge for each extra day is ₹3.

Page No 150:

Question 48:

A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each
should be used to make 10 litres of a 40% acid solution?

Answer:

Let x litres and y litres be the amount of acids from 50% and 25% acid solutions respectively.
As per the question
$50 % of x + 25 % of y = 40 % of 10 \Rightarrow 0.50 x + 0.25 y = 4 \Rightarrow 2 x + y = 16 . . . . . (i)$
Since, the total volume is 10 litres, so
$x + y = 10 . . . . . (i i)$
Subtracting (ii) from (i), we get
$x = 6$
Now, putting x = 6 in (ii), we have
$6 + y = 10 \Rightarrow y = 4$
Hence, volume of 50% acid solution = 6 litres and volume of 25% acid solution = 4 litres.

Page No 150:

Question 49:

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a
bar of 16-carat gold, weighing 120 g?

Answer:

Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition
$\frac{18 x}{24} + \frac{12 y}{24} = \frac{120 \times 16}{24} \Rightarrow 3 x + 2 y = 320 . . . . . (i)$
And
$x + y = 120 . . . . . (i i)$
Multiplying (ii) by 2 and subtracting from (i), we get
$x = 320 - 240 = 80$
Now, putting x = 80 in (ii), we have
$80 + y = 120 \Rightarrow y = 40$
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.

Page No 150:

Question 50:

90% and 97% pure acid solutions are mixed to obtain 21 litres of 95% pure acid solution.
Find the quantity of each type of acids to be mixed to form the mixture.

Answer:

Let x litres and y litres be respectively the amount of 90% and 97% pure acid solutions.
As per the given condition
$0.90 x + 0.97 y = 21 \times 0.95 \Rightarrow 0.90 x + 0.97 y = 21 \times 0.95 . . . . . (i)$
And
$x + y = 21 . . . . . (i i)$
From (ii), subtitute $y = 21 - x$ in (i) to get
$0.90 x + 0.97 (21 - x) = 21 \times 0.95 \Rightarrow 0.90 x + 0.97 \times 21 - 0.97 x = 21 \times 0.95 \Rightarrow 0.07 x = 0.97 \times 21 - 21 \times 0.95 \Rightarrow x = \frac{21 \times 0.02}{0.07} = 6$
Now, putting x = 6 in (ii), we have
$6 + y = 21 \Rightarrow y = 15$
Hence, the required quantities are 6 litres and 15 litres.

Page No 150:

Question 51:

The larger of the two supplementary angles exceeds the smaller by $18^{\circ}$ . Find them.

Answer:

Let x and y be the supplementary angles, where x > y.
As per the given condition
$x + y = 180^{\circ} . . . . . (i)$
And
$x - y = 18^{\circ} . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 198^{\circ} \Rightarrow x = 99^{\circ}$
Now, substituting $x = 99^{\circ}$ in (ii), we have
$99^{\circ} - y = 18^{\circ} \Rightarrow x = 99^{\circ} - 18^{\circ} = 81^{\circ}$
Hence, the required angles are $99^{\circ}$ and $81^{\circ}$ .

Page No 150:

Question 52:

In $∆ A B C, ∠ A = x^{\circ}, ∠ B = {(3 x - 2)}^{\circ}, ∠ C = y^{\circ} and ∠ C - ∠ B = 9^{\circ} .$ Find the three angles.

Answer:

$∵ ∠ C - ∠ B = 9^{\circ} ∴ y^{\circ} - {(3 x - 2)}^{\circ} = 9^{\circ} \Rightarrow y^{\circ} - 3 x^{\circ} + 2^{\circ} = 9^{\circ} \Rightarrow y^{\circ} - 3 x^{\circ} = 7^{\circ} . . . . . (i)$
The sum of all the angles of a triangle is $180^{\circ}$ , therefore
$∠ A + ∠ B + ∠ C = 180^{\circ} \Rightarrow x^{\circ} + {(3 x - 2)}^{\circ} + y^{\circ} = 180^{\circ} \Rightarrow 4 x^{\circ} + y^{\circ} = 182^{\circ} . . . . . (i i)$
Subtracting (i) from (ii), we have
$7 x^{\circ} = 182^{\circ} - 7^{\circ} = 175^{\circ} \Rightarrow x^{\circ} = 25^{\circ}$
Now, substituting $x^{\circ} = 25^{\circ}$ in (i), we have
$y^{\circ} = 3 x^{\circ} + 7^{\circ} = 3 \times 25^{\circ} + 7^{\circ} = 82^{\circ}$
Thus
$∠ A = x^{\circ} = 25^{\circ} ∠ B = {(3 x - 2)}^{\circ} = 75^{\circ} - 2^{\circ} = 73^{\circ} ∠ C = y^{\circ} = 82^{\circ}$
Hence, the angles are $25^{\circ}, 73^{\circ} and 82^{\circ}$ .

Page No 150:

Question 53:

In a cyclic quadrilateral ABCD, it is given that $∠ A = {(2 x + 4)}^{\circ}, ∠ B = {(y + 3)}^{\circ}, ∠ C = {(2 y + 10)}^{\circ} and ∠ D = {(4 x - 5)}^{\circ}$ .
Find the four angles.

Answer:

The opposite angles of cyclic quadrilateral are supplementary, so
$∠ A + ∠ C = 180^{\circ} \Rightarrow {(2 x + 4)}^{\circ} + {(2 y + 10)}^{\circ} = 180^{\circ} \Rightarrow x + y = 83^{\circ} . . . . . (i)$
And
$∠ B + ∠ D = 180^{\circ} \Rightarrow {(y + 3)}^{\circ} + {(4 x - 5)}^{\circ} = 180^{\circ} \Rightarrow 4 x + y = 182^{\circ} . . . . . (i i)$
Subtracting (i) from (ii), we have
$3 x = 99 \Rightarrow x = 33^{\circ}$
Now, substituting $x = 33^{\circ}$ in (i), we have
$33^{\circ} + y = 83^{\circ} \Rightarrow y = 83^{\circ} - 33^{\circ} = 50^{\circ}$
Therefore
$∠ A = {(2 x + 4)}^{\circ} = {(2 \times 33 + 4)}^{\circ} = 70^{\circ} ∠ B = {(y + 3)}^{\circ} = {(50 + 3)}^{\circ} = 53^{\circ} ∠ C = {(2 y + 10)}^{\circ} = {(2 \times 50 + 10)}^{\circ} = 110^{\circ} ∠ D = {(4 x - 5)}^{\circ} = {(4 \times 33 - 5)}^{\circ} = 132 ° - 5^{\circ} = 127^{\circ}$
Hence, $∠ A = 70^{\circ}, ∠ B = 53^{\circ}, ∠ C = 110^{\circ} and ∠ D = 127^{\circ}$ .

Page No 155:

Question 1:

Write the number of solutions of the following pair of linear equations:
$x + 2 y - 8 = 0, 2 x + 4 y = 16$

Answer:

The given equations are
$x + 2 y - 8 = 0 . . . . . (i) 2 x + 4 y - 16 = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 1, b_{1} = 2, c_{1} = - 8, a_{2} = 2, b_{2} = 4 and c_{2} = - 18$
Now
$\frac{a_{1}}{a_{2}} = \frac{1}{2} \frac{b_{1}}{b_{2}} = \frac{2}{4} = \frac{1}{2} \frac{c_{1}}{c_{2}} = \frac{- 8}{- 16} = \frac{1}{2}$
$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} = \frac{1}{2}$
Thus, the pair of linear equations are coincident and therefore has infinitely many solutions.

Page No 155:

Question 2:

Find the value of k for which the following pair of linear equations have infinitely many solutions:
$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k$

Answer:

The given equations are
$2 x + 3 y - 7 = 0 . . . . . (i) (k - 1) x + (k + 2) y - 3 k = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 2, b_{1} = 3, c_{1} = - 7, a_{2} = k - 1, b_{2} = k + 2 and c_{2} = - 3 k$
For the given pair of linear equations to have infinitely many solutions, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{- 7}{- 3 k} \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2}, \frac{3}{k + 2} = \frac{- 7}{- 3 k} and \frac{2}{k - 1} = \frac{- 7}{- 3 k} \Rightarrow 2 (k + 2) = 3 (k - 1), 9 k = 7 k + 14 and 6 k = 7 k - 7$
$\Rightarrow k = 7, k = 7 and k = 7$
Hence, k = 7.

Page No 155:

Question 3:

For what value of k does the following pair of linear equations have infinitely many solutions?:
$10 x + 5 y - (k - 5) = 0, 20 x + 10 y - k = 0$

Answer:

The given pair of linear equations are
$10 x + 5 y - (k - 5) = 0 . . . . . (i) 20 x + 10 y - k = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 10, b_{1} = 5, c_{1} = - (k - 5), a_{2} = 20, b_{2} = 10 and c_{2} = - k$
For the given pair of linear equations to have infinitely many solutions, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{10}{20} = \frac{5}{10} = \frac{- (k - 5)}{- k} \Rightarrow \frac{1}{2} = \frac{k - 5}{k} \Rightarrow 2 k - 10 = k \Rightarrow k = 10$
Hence, k = 10.

Page No 155:

Question 4:

For what value of k will the following pair of linear equations have no solution?:
$2 x + 3 y = 9, 6 x + (k - 2) y = (3 k - 2)$

Answer:

The given pair of linear equations are
$2 x + 3 y - 9 = 0 . . . . . (i) 6 x + (k - 2) y - (3 k - 2) = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 2, b_{1} = 3, c_{1} = - 9, a_{2} = 6, b_{2} = k - 2 and c_{2} = - (3 k - 2)$
For the given pair of linear equations to have no solution, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{6} = \frac{3}{k - 2} \neq \frac{- 9}{- (3 k - 2)} \Rightarrow \frac{2}{6} = \frac{3}{k - 2}, \frac{3}{k - 2} \neq \frac{- 9}{- (3 k - 2)} \Rightarrow k = 11, \frac{3}{k - 2} \neq \frac{9}{3 k - 2}$
$\Rightarrow k = 11, 3 (3 k - 2) \neq 9 (k - 2) \Rightarrow k = 11, 1 \neq 3 (true)$
Hence, k = 11.

Page No 155:

Question 5:

Write the number of solutions of the following pair of linear equations:
$x + 3 y - 4 = 0, 2 x + 6 y - 7 = 0$

Answer:

The given pair of linear equations are
$x + 3 y - 4 = 0 . . . . . (i) 2 x + 6 y - 7 = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 1, b_{1} = 3, c_{1} = - 4, a_{2} = 2, b_{2} = 6 and c_{2} = - 7$
Now
$\frac{a_{1}}{a_{2}} = \frac{1}{2} \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} \frac{c_{1}}{c_{2}} = \frac{- 4}{- 7} = \frac{4}{7} \Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Thus, the pair of the given linear equations has no solution.

Page No 155:

Question 6:

Write the value of k for which the system of equations $3 x + k y = 0, 2 x - y = 0$ has a unique solution.

Answer:

The given pair of linear equations is
$3 x + k y = 0 . . . . . (i) 2 x - y = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 3, b_{1} = k, c_{1} = 0, a_{2} = 2, b_{2} = - 1 and c_{2} = 0$
For the system to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \Rightarrow \frac{3}{2} \neq \frac{k}{- 1} \Rightarrow k \neq - \frac{3}{2}$
Hence, $k \neq - \frac{3}{2}$ .

Page No 155:

Question 7:

The difference between two numbers is 5 and the difference between their squares is 65. Find the numbers.

Answer:

Let the numbers be x and y, where x > y.
Then as per the question
$x - y = 5 . . . . . (i) x^{2} - y^{2} = 65 . . . . . (i i)$
Dividing (ii) by (i), we get
$\frac{x^{2} - y^{2}}{x - y} = \frac{65}{5} \Rightarrow \frac{(x - y) (x + y)}{x - y} = 13 \Rightarrow x + y = 13 . . . . . (i i i)$
Now, adding (i) and (ii), we have
$2 x = 18 \Rightarrow x = 9$
Substituting x = 9 in (iii), we have
$9 + y = 13 \Rightarrow y = 4$
Hence, the numbers are 9 and 4.

Page No 155:

Question 8:

The cost of 5 pens and 8 pencils is ₹120, while the cost of 8 pens and 5 pencils is ₹153.
Find the cost of 1 pen and that of 1 pencil.

Answer:

Let the cost of 1 pen and 1 pencil are ₹x and ₹y respectively.
Then as per the question
$5 x + 8 y = 120 . . . . . (i) 8 x + 5 y = 153 . . . . . (i i)$
Adding (i) and (ii), we get
$13 x + 13 y = 273 \Rightarrow x + y = 21 . . . . . (i i i)$
Subtracting (i) from (ii), we get
$3 x - 3 y = 33 \Rightarrow x - y = 11 . . . . . (i v)$
Now, adding (iii) and (iv), we get
$2 x = 32 \Rightarrow x = 16$
Substituting x = 16 in (iii), we have
$16 + y = 21 \Rightarrow y = 5$
Hence, the cost of 1 pen and 1 pencil are respectively ₹16 and ₹5.

Page No 155:

Question 9:

The sum of two numbers is 80. The larger number exceeds four times the
smaller one by 5. Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then as per the question
$x + y = 80 . . . . . (i) x = 4 y + 5 x - 4 y = 5 . . . . . (i i)$
Subtracting (ii) from (i), we get
$5 y = 75 \Rightarrow y = 15$
Now, putting y = 15 in (i), we have
$x + 15 = 80 \Rightarrow x = 65$
Hence, the numbers are 65 and 15.

Page No 155:

Question 10:

A number consists of two digits whose sum is 10. If 18 is subtracted from the number, its
digits are reversed. Find the number.

Answer:

Let the ones digit and tens digit be x and y respectively.
Then as per the question
$x + y = 10 . . . . . (i) (10 y + x) - 18 = 10 x + y x - y = - 2 . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 8 \Rightarrow x = 4$
Now, putting x = 4 in (i), we have
$4 + y = 10 \Rightarrow y = 6$
Hence, the number is 64.

Page No 155:

Question 11:

A man purchased 47 stamps of 20 p and 25 p for ₹10. Find the number of each type of stamps.

Answer:

Let the number of stamps of 20 p and 25 p be x and y respectively.
As per the question
$x + y = 47 . . . . . (i) 0.20 x + 0.25 y = 10 4 x + 5 y = 200 . . . . . (i i)$
From (i), we get
$y = 47 - x$
Now, substituting $y = 47 - x$ in (ii), we have
$4 x + 5 (47 - x) = 200 \Rightarrow 4 x - 5 x + 235 = 200 \Rightarrow x = 235 - 200 = 35$
Putting x = 35 in (i), we get
$35 + y = 47 \Rightarrow y = 47 - 35 = 12$
Hence, the number of 20 p stamps and 25 p stamps are 35 and 12 respectively.

Page No 155:

Question 12:

A man has some hens and cows. If the number of heads be 48 and the number of feet
be 140, how many cows are there?

Answer:

Let the number of hens and cow be x and y respectively.
As per the question
$x + y = 48 . . . . . (i) 2 x + 4 y = 140 x + 2 y = 70 . . . . . (i i)$
Subtracting (i) from (ii), we have
$y = 22$
Hence, the number of cows is 22.

Page No 155:

Question 13:

If $\frac{2}{x} + \frac{3}{y} = \frac{9}{x y} and \frac{4}{x} + \frac{9}{y} = \frac{21}{x y}$ , find the values of x and y.

Answer:

The given pair of equation is
$\frac{2}{x} + \frac{3}{y} = \frac{9}{x y} . . . . . (i) \frac{4}{x} + \frac{9}{y} = \frac{21}{x y} . . . . . (i i)$
Multiplying (i) and (ii) by xy, we have
$3 x + 2 y = 9 . . . . . (i i i) 9 x + 4 y = 21 . . . . . (i v)$
Now, multiplying (iii) by 2 and subtracting from (iv), we get
$9 x - 6 x = 21 - 18 \Rightarrow x = \frac{3}{3} = 1$
Putting x = 1 in (iii), we have
$3 \times 1 + 2 y = 9 \Rightarrow y = \frac{9 - 3}{2} = 3$
Hence, x = 1 and y = 3.

Page No 155:

Question 14:

If $\frac{x}{4} + \frac{y}{3} = \frac{5}{12} and \frac{x}{2} + y = 1$ , then find the value of (x + y).

Answer:

The given pair of equations is
$\frac{x}{4} + \frac{y}{3} = \frac{5}{12} . . . . . (i) \frac{x}{2} + y = 1 . . . . . (i i)$
Multiplying (i) by 12 and (ii) by 4, we get
$3 x + 4 y = 5 . . . . . (i i i) 2 x + 4 y = 4 . . . . . (i v)$
Now, subtracting (iv) from (iii), we get
$x = 1$
Putting x = 1 in (iv), we have
$2 + 4 y = 4 \Rightarrow 4 y = 2 \Rightarrow y = \frac{1}{2}$
$∴ x + y = 1 + \frac{1}{2} = \frac{3}{2}$
Hence, the value of x + y is $\frac{3}{2}$ .

Page No 155:

Question 15:

If $12 x + 17 y = 53 and 17 x + 12 y = 63$ , then find the value of (x + y).

Answer:

The given pair of equations is
$12 x + 17 y = 53 . . . . . (i) 17 x + 12 y = 63 . . . . . (i i)$
Adding (i) and (ii), we get
$29 x + 29 y = 116 \Rightarrow x + y = 4 (Dividing by 4)$
Hence, the value of x + y is 4.

Page No 156:

Question 16:

Find the value of k for which the system 3x + 5y = 0 and kx + 10y = 0 has a nonzero solution.

Answer:

The given system is
$3 x + 5 y = 0 . . . . . (i) k x + 10 y = 0 . . . . . (i i)$
This is a homogeneous system of linear differential equation, so it always has a zero
solution i.e., x = y = 0.
But to have a nonzero solution, it must have infinitely many solutions.
For this, we have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \Rightarrow \frac{3}{k} = \frac{5}{10} = \frac{1}{2} \Rightarrow k = 6$
Hence, k = 6.

Page No 156:

Question 17:

Find k for which the system kx − y = 2 and 6x − 2y = 3 has a unique solution.

Answer:

The given system is
$k x - y - 2 = 0 . . . . . (i) 6 x - 2 y - 3 = 0 . . . . . (i i)$
Here, $a_{1} = k, b_{1} = - 1, c_{1} = - 2 . a_{2} = 6, b_{2} = - 2 and c_{2} = - 3$ .
For the system, to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{k}{6} \neq \frac{- 1}{- 2} = \frac{1}{2} \Rightarrow k \neq 3$
Hence, $k \neq 3$ .

Page No 156:

Question 18:

Find k for which the system 2x + 3y − 5 = 0 and 4x + ky − 10 = 0 has an infinite number of solutions.

Answer:

The given system is
$2 x + 3 y - 5 = 0 . . . . . (i) 4 x + k y - 10 = 0 . . . . . (i i)$
Here, $a_{1} = 2, b_{1} = 3, c_{1} = - 5 . a_{2} = 4, b_{2} = k and c_{2} = - 10$ .
For the system, to have an infinite number of solutions, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{4} = \frac{3}{k} = \frac{- 5}{- 10} \Rightarrow \frac{1}{2} = \frac{3}{k} = \frac{1}{2} \Rightarrow k = 6$
Hence, k = 6.

Page No 156:

Question 19:

Show that the system 2x + 3y − 1 = 0, 4x + 6y − 4 = 0 has no solution.

Answer:

The given system is
$2 x + 3 y - 1 = 0 . . . . . (i) 4 x + 6 y - 4 = 0 . . . . . (i i)$
Here, $a_{1} = 2, b_{1} = 3, c_{1} = - 1 . a_{2} = 4, b_{2} = 6 and c_{2} = - 4$ .
Now,
$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2} \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} \frac{c_{1}}{c_{2}} = \frac{- 1}{- 4} = \frac{1}{4}$
Thus, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ and therefor the given system has no solution.

Page No 156:

Question 20:

Find k for which the system x + 2y = 3 and 5x + ky + 7 = 0 is inconsistent.

Answer:

The given system is
$x + 2 y - 3 = 0 . . . . . (i) 5 x + k y + 7 = 0 . . . . . (i i)$
Here, $a_{1} = 1, b_{1} = 2, c_{1} = - 3, a_{2} = 5, b_{2} = k and c_{2} = 7$ .
For the system to be inconsistent, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{1}{5} = \frac{2}{k} \neq \frac{- 3}{7} \Rightarrow \frac{1}{5} = \frac{2}{k} \Rightarrow k = 10$
Hence, k = 10.

Page No 156:

Question 21:

Solve: $\frac{3}{x + y} + \frac{2}{x - y} = 2 and \frac{9}{x + y} - \frac{4}{x - y} = 1$ .

Answer:

The given system of equations is
$\frac{3}{x + y} + \frac{2}{x - y} = 2 . . . . . (i) \frac{9}{x + y} - \frac{4}{x - y} = 1 . . . . . (i i)$
Substituting $\frac{1}{x + y} = u and \frac{1}{x - y} = v$ in (i) and (ii), the given equations are changed to
$3 u + 2 v = 2 . . . . . (i i i) 9 u - 4 v = 1 . . . . . (i v)$
Multiplying (i) by 2 and adding it with (ii), we get
$15 u = 4 + 1 \Rightarrow u = \frac{1}{3}$
Multiplying (i) by 3 and subtracting (ii) from it, we get
$6 v + 4 v = 6 - 1 \Rightarrow u = \frac{5}{10} = \frac{1}{2}$
Therefore
$x + y = 3 . . . . . (v) x - y = 2 . . . . . (v i)$
Now, adding (v) and (vi) we have
$2 x = 5 \Rightarrow x = \frac{5}{2}$
Substituting $x = \frac{5}{2}$ in (v), we have
$\frac{5}{2} + y = 3 \Rightarrow y = 3 - \frac{5}{2} = \frac{1}{2}$
Hence, $x = \frac{5}{2} and y = \frac{1}{2}$ .

Page No 158:

Question 1:

If 2x + 3y = 12 and 3x − 2y = 5 then
(a) x = 2, y = 3 (b) x = 2, y = − 3 (c) x = 3, y = 2 (d) x = 3, y = − 2

Answer:

The given system of equations is
$2 x + 3 y = 12 . . . . . (i) 3 x - 2 y = 5 . . . . . (i i)$
Multiplying (i) by 2 and (ii) by 3 and then adding, we get
$4 x + 9 x = 24 + 15 \Rightarrow x = \frac{39}{13} = 3$
Now, putting x = 3 in (i), we have
$2 \times 3 + 3 y = 12 \Rightarrow y = \frac{12 - 6}{3} = 2$
Thus, x = 3 and y = 2.
Hence, the correct answer is option (c).

Page No 158:

Question 2:

If $x - y = 2 and \frac{2}{x + y} = \frac{1}{5}$ then
(a) x = 4, y = 2 (b) x = 5, y = 3 (c) x = 6, y = 4 (d) x = 7, y = 5

Answer:

The given system of equations is
$x - y = 2 . . . . . (i) x + y = 10 . . . . . (i i)$
Adding (i) and (ii), we get
$2 x = 12 \Rightarrow x = 6$
Now, putting x = 6 in (ii), we have
$6 + y = 10 \Rightarrow y = 10 - 6 = 4$
Thus, x = 6 and y = 4.
Hence, the correct answer is option (c).

Page No 158:

Question 3:

If $\frac{2 x}{3} - \frac{y}{2} + \frac{1}{6} = 0 and \frac{x}{2} + \frac{2 y}{3} = 3$ then
(a) x = 2, y = 3 (b) x = − 2, y = 3 (c) x = 2, y = − 3 (d) x = − 2, y = − 3

Answer:

The given system of equations is
$\frac{2 x}{3} - \frac{y}{2} = - \frac{1}{6} . . . . . (i) \frac{x}{2} + \frac{2 y}{3} = 3 . . . . . (i i)$
Multiplying (i) and (ii) by 6, we get
$4 x - 3 y = - 1 . . . . . (i i i) 3 x + 4 y = 18 . . . . . (i v)$
Multiplying (iii) by 4 and (iv) by 3 and adding, we get
$16 x + 9 x = - 4 + 54 \Rightarrow x = \frac{50}{25} = 2$
Now, putting x = 2 in (iv), we have
$3 \times 2 + 4 y = 18 \Rightarrow y = \frac{18 - 6}{4} = 3$
Thus, x = 2 and y = 3.
Hence, the correct answer is option (a).

Page No 158:

Question 4:

If $\frac{1}{x} + \frac{2}{y} = 4 and \frac{3}{y} - \frac{1}{x} = 11$ then
(a) x = 2, y = 3 (b) x = − 2, y = 3 (c) $x = - \frac{1}{2}, y = 3$ (d) $x = - \frac{1}{2}, y = \frac{1}{3}$

Answer:

The given system of equations is
$\frac{1}{x} + \frac{2}{y} = 4 . . . . . (i) \frac{3}{y} - \frac{1}{x} = 11 . . . . . (i i)$
Adding (i) and (ii), we get
$\frac{2}{y} + \frac{3}{y} = 15 \Rightarrow \frac{5}{y} = 15 \Rightarrow y = \frac{5}{15} = \frac{1}{3}$
Now, putting $y = \frac{1}{3}$ in (i), we have
$\frac{1}{x} + 2 \times 3 = 4 \Rightarrow \frac{1}{x} = 4 - 6 \Rightarrow x = - \frac{1}{2}$
Thus, $x = - \frac{1}{2}, y = \frac{1}{3}$ .
Hence, the correct answer is option (d).

Page No 158:

Question 5:

If $\frac{2 x + y + 2}{5} = \frac{3 x - y + 1}{3} = \frac{3 x + 2 y + 1}{6}$ then
(a) x = 1, y = 1 (b) x = − 1, y = − 1 (c) x = 1, y = 2 (d) x = 2, y = 1

Answer:

Consider $\frac{2 x + y + 2}{5} = \frac{3 x - y + 1}{3} and \frac{3 x - y + 1}{3} = \frac{3 x + 2 y + 1}{6}$ . Now, simplifying these equations, we get
$3 (2 x + y + 2) = 5 (3 x - y + 1) \Rightarrow 6 x + 3 y + 6 = 15 x - 5 y + 5 \Rightarrow 9 x - 8 y = 1 . . . . . (i)$
And
$6 (3 x - y + 1) = 3 (3 x + 2 y + 1) \Rightarrow 18 x - 6 y + 6 = 9 x + 6 y + 3 \Rightarrow 3 x - 4 y = - 1 . . . . . (i i)$
Multiplying (ii) by 2 and subtracting it from (i)
$9 x - 6 x = 1 + 2 \Rightarrow x = 1$
Now, putting x = 1 in (ii), we have
$3 \times 1 - 4 y = - 1 \Rightarrow y = \frac{3 + 1}{4} = 1$
Thus, x = 1, y = 1.
Hence, the correct answer is option (a).

Page No 158:

Question 6:

If $\frac{3}{x + y} + \frac{2}{x - y} = 2 and \frac{9}{x + y} - \frac{4}{x - y} = 1$ then
(a) $x = \frac{1}{2}, y = \frac{3}{2}$ (b) $x = \frac{5}{2}, y = \frac{1}{2}$ (c) $x = \frac{3}{2}, y = \frac{1}{2}$ (d) $x = \frac{1}{2}, y = \frac{5}{2}$

Answer:

The given equations are
$\frac{3}{x + y} + \frac{2}{x - y} = 2 . . . . . (i) \frac{9}{x + y} - \frac{4}{x - y} = 1 . . . . . (i i)$
Substituting $\frac{1}{x + y} = u and \frac{1}{x - y} = v$ in (i) and (ii), the new system becomes
$3 u + 2 v = 2 . . . . . (i i i) 9 u - 4 v = 1 . . . . . (i v)$
Now, multiplying (iii) by 2 and adding it with (iv), we get
$6 u + 9 u = 4 + 1 \Rightarrow u = \frac{5}{15} = \frac{1}{3}$
Again, multiplying (iii) by 3 and subtracting (iv) from it, we get
$6 v + 4 v = 6 - 1 \Rightarrow v = \frac{5}{10} = \frac{1}{2}$
Therefore
$x + y = 3 . . . . . (v) x - y = 2 . . . . . (v i)$
Adding (v) and (vi), we get
$2 x = 3 + 2 \Rightarrow x = \frac{5}{2}$
Substituting $x = \frac{5}{2}$ , in (v), we have
$\frac{5}{2} + y = 3 \Rightarrow y = 3 - \frac{5}{2} = \frac{1}{2}$
Thus, $x = \frac{5}{2} and y = \frac{1}{2}$ .
Hence, the correct answer is option (b).

Page No 158:

Question 7:

If $4 x + 6 y = 3 x y and 8 x + 9 y = 5 x y$ then
(a) x = 2, y = 3 (b) x = 1, y = 2 (c) x = 3, y = 4 (d) x = 1, y = − 1

Answer:

The given equations are
$4 x + 6 y = 3 x y . . . . . (i) 8 x + 9 y = 5 x y . . . . . (i i)$
Dividing (i) and (ii) by xy, we get
$\frac{6}{x} + \frac{4}{y} = 3 . . . . . (i i i) \frac{9}{x} + \frac{8}{y} = 5 . . . . . (i v)$
Multiplying (iii) by 2 and subtracting (iv) from it, we get
$\frac{12}{x} - \frac{9}{x} = 6 - 5 \Rightarrow \frac{3}{x} = 1 \Rightarrow x = 3$
Substituting x = 3 in (iii), we get
$\frac{6}{3} + \frac{4}{y} = 3 \Rightarrow \frac{4}{y} = 1 \Rightarrow y = 4$
Thus, x = 3 and y = 4.
Hence, the correct answer is option (c).

Page No 158:

Question 8:

If $29 x + 37 y = 103 and 37 x + 29 y = 95$ then
(a) x = 1, y = 2 (b) x = 2, y = 1 (c) x = 3, y = 2 (d) x = 2, y = 3

Answer:

The given equations are
$29 x + 37 y = 103 . . . . . (i) 37 x + 29 y = 95 . . . . . (i i)$
Adding (i) and (ii), we get
$66 x + 66 y = 198 \Rightarrow x + y = 3 . . . . . (i i i)$
Subtracting (i) from (ii), we get
$8 x - 8 y = - 8 \Rightarrow x - y = - 1 . . . . . (i v)$
Adding (iii) and (iv), we get
$2 x = 2 \Rightarrow x = 1$
Substituting x = 1 in (iii), we have
$1 + y = 3 \Rightarrow y = 2$
Thus, x = 1 and y = 2.
Hence, the correct answer is option (a).

Page No 158:

Question 9:

If $2^{x + y} = 2^{x - y} = \sqrt{8}$ then the value of y is
(a) $\frac{1}{2}$ (b) $\frac{3}{2}$ (c) 0 (d) none of these

Answer:

$∵ 2^{x + y} = 2^{x - y} = \sqrt{8} ∴ x + y = x - y \Rightarrow y = 0$
Hence, the correct answer is option (c).

Page No 159:

Question 10:

If $\frac{2}{x} + \frac{3}{y} = 6 and \frac{1}{x} + \frac{1}{2 y} = 2$ then
(a) $x = 1, y = \frac{2}{3}$ (b) $x = \frac{2}{3}, y = 1$ (c) $x = 1, y = \frac{3}{2}$ (d) $x = \frac{3}{2}, y = 1$

Answer:

The given equations are
$\frac{2}{x} + \frac{3}{y} = 6 . . . . . (i) \frac{1}{x} + \frac{1}{2 y} = 2 . . . . . (i i)$
Multiplying (ii) by 2 and subtracting it from (ii), we get
$\frac{3}{y} - \frac{1}{y} = 6 - 4 \Rightarrow \frac{2}{y} = 2 \Rightarrow y = 1$
Substituting y = 1 in (ii), we get
$\frac{1}{x} + \frac{1}{2} = 2 \Rightarrow \frac{1}{x} = 2 - \frac{1}{2} = \frac{3}{2} \Rightarrow x = \frac{2}{3}$
Hence, the correct answer is option (b).

Page No 159:

Question 11:

The system $k x - y = 2 and 6 x - 2 y = 3$ has a unique solution only when
(a) k = 0 (b) $k \neq 0$ (c) k = 3 (d) $k \neq 3$

Answer:

The given equations are
$k x - y - 2 = 0 . . . . . (i) 6 x - 2 y - 3 = 0 . . . . . (i i)$
Here, $a_{1} = k, b_{1} = - 1, c_{1} = - 2, a_{2} = 6, b_{2} = - 2 and c_{2} = - 3 .$
For the given system to have a unique solution, we must have
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{k}{6} \neq \frac{- 1}{- 2} \Rightarrow k \neq 3$
Hence, the correct answer is option (d).

Page No 159:

Question 12:

The system $x - 2 y = 3 and 3 x + k y = 1$ has a unique solution only when

(a) k = −6
(b) k ≠ −6
(c) k = 0
(d) k ≠ 0

Answer:

The correct option is (b).

The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = −2, c₁ = −3 and a₂ = 3, b₂ = k and c₂ = −1
∴ $\frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{k} and \frac{c_{1}}{c_{2}} = \frac{- 3}{- 1} = 3$
These graph lines will intersect at a unique point when we have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ ⇒ $\frac{1}{3} \neq \frac{- 2}{k} \Rightarrow k \neq - 6$
Hence, k has all real values other than −6.

Page No 159:

Question 13:

The system $x + 2 y = 3 and 5 x + k y + 7 = 0$ has no solution, when

(a) k = 10
(b) $k \neq 10$
(c) $k = \frac{- 7}{3}$
(d) k = −21

Answer:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 5x + ky + 7 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = 2, c₁ = −3 and a₂ = 5, b₂ = k and c₂ = 7
∴ $\frac{a_{1}}{a_{2}} = \frac{1}{5}, \frac{b_{1}}{b_{2}} = \frac{2}{k} and \frac{c_{1}}{c_{2}} = \frac{- 3}{7}$
For the system of equations to have no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ $\frac{1}{5} = \frac{2}{k} \neq \frac{- 3}{7} \Rightarrow k = 10$

Page No 159:

Question 14:

If the lines given by $3 x + 2 k y = 2 and 2 x + 5 y + 1 = 0$ are parallel, then the value of k is

(a) $\frac{- 5}{4}$
(b) $\frac{2}{5}$
(c) $\frac{3}{2}$
(d) $\frac{15}{4}$

Answer:

The correct option is (d).

The given system of equations can be written as follows:
3x + 2ky − 2 = 0 and 2x + 5y + 1 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁ = 2k, c₁ = −2 and a₂ = 2, b₂ = 5 and c₂ = 1
∴ $\frac{a_{1}}{a_{2}} = \frac{3}{2}, \frac{b_{1}}{b_{2}} = \frac{2 k}{5} and \frac{c_{1}}{c_{2}} = \frac{- 2}{1}$
For parallel lines, we have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
∴ $\frac{3}{2} = \frac{2 k}{5} \neq \frac{- 2}{1}$
⇒ $k = \frac{15}{4}$

Page No 159:

Question 15:

For what value of k do the equations $k x - 2 y = 3 and 3 x + y = 5$ represent two lines intersecting at a unique point?

(a) k = 3
(b) k = −3
(c) k = 6
(d) all real values except −6

Answer:

The correct option is (d).

The given system of equations can be written as follows:
kx − 2y − 3 = 0 and 3x + y − 5 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁ = −2, c₁ = −3 and a₂ = 3, b₂ = 1 and c₂ = −5
∴ $\frac{a_{1}}{a_{2}} = \frac{k}{3}, \frac{b_{1}}{b_{2}} = \frac{- 2}{1} and \frac{c_{1}}{c_{2}} = \frac{- 3}{- 5} = \frac{3}{5}$
Thus, for these graph lines to intersect at a unique point, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
⇒ $\frac{k}{3} \neq \frac{- 2}{1} \Rightarrow k \neq - 6$
Hence, the graph lines will intersect at all real values of k except −6.

Page No 159:

Question 16:

The pair of equations $x + 2 y + 5 = 0 and - 3 x - 6 y + 1 = 0$ has

(a) a unique solutions
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

Answer:

The correct option is (d).

The given system of equations can be written as:
x + 2y + 5 = 0 and −3x − 6y + 1 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = 2, c₁ = 5 and a₂ = −3, b₂ = −6 and c₂ = 1
∴ $\frac{a_{1}}{a_{2}} = \frac{1}{- 3}, \frac{b_{1}}{b_{2}} = \frac{2}{- 6} = \frac{1}{- 3} and \frac{c_{1}}{c_{2}} = \frac{5}{1}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the given system has no solution.

Page No 159:

Question 17:

The pair of equations $2 x + 3 y = 5 and 4 x + 6 y = 15$ has

(a) a unique solution
(b) exactly two solutions
(c) infinitely many solutions
(d) no solution

Answer:

The correct option is (d).

The given system of equations can be written as follows:
2x + 3y − 5 = 0 and 4x + 6y − 15 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = 3, c₁ = −5 and a₂ = 4, b₂ = 6 and c₂ = −15
∴ $\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2} and \frac{c_{1}}{c_{2}} = \frac{- 5}{- 15} = \frac{1}{3}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the given system has no solution.

Page No 159:

Question 18:

If a pair of linear equations is consistent, then their graph lines will be

(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident

Answer:

The correct option is (d).

If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincide.

Page No 159:

Question 19:

If a pair of linear equations is inconsistent, then their graph lines will be

(a) parallel
(b) always coincident
(c) always intersecting
(d) intersecting or coincident

Answer:

The correct option is (a).

If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their graph lines do not intersect. Hence, they are parallel.

Page No 159:

Question 20:

In a $∆ A B C, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B), then ∠ B = ?$
(a) 20°
(b) 40°
(c) 60°
(d) 80°

Answer:

The correct option is (b).

Let $∠ A = x ° a n d ∠ B = y °$
∴ $∠ C = 3 ∠ B = (3 y) °$
Now, $∠ A + ∠ B + ∠ C = 180 °$
⇒ x + y + 3y = 180
⇒ x + 4y = 180              ...(i)
Also, $∠ C = 2 (∠ A + ∠ B)$
⇒ 3y = 2(x + y)
⇒ 2x − y = 0                  ...(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0                    ...(iii)
On adding (i) and (iii) we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴ $∠ B = y ° = 40 °$

Page No 159:

Question 21:

In a cyclic quadrilateral ABCD, it is being given that $∠ A = (x + y + 10) °, ∠ B = (y + 20) °, ∠ C = (x + y - 30) ° and ∠ D = (x + y) ° . Then, ∠ B = ?$
(a) 70°
(b) 80°
(c) 100°
(d) 110°

Answer:

The correct option is (b).
In a cyclic quadrilateral ABCD:
$∠ A = (x + y + 10) °$
$∠ B = (y + 20 °)$
$∠ C = (x + y - 30) °$
$∠ D = (x + y) °$
We have:
$∠ A + ∠ C = 180 °$ and $∠ B + ∠ D = 180 °$       [Since ABCD is a cyclic quadrilateral]
Now, $∠ A + ∠ C = (x + y + 10) ° + (x + y - 30) ° = 180 °$
⇒ 2x + 2y − 20 = 180
⇒ x + y − 10 = 90
⇒ x + y = 100                   ....(i)
Also, $∠ B + ∠ D = (y + 20) ° + (x + y) ° = 180 °$
⇒ x + 2y + 20 = 180
⇒ x + 2y = 160                ....(ii)
On subtracting (i) from (ii), we get:
y = (160 − 100) = 60
On substituting y = 60 in (i), we get:
x + 60 = 100 ⇒ x = (100 − 60) = 40
∴ $∠ B = (y + 20 °) = (60 + 20) ° = 80 °$

Page No 160:

Question 22:

The sum of the digits of a two digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. The number is

(a) 96
(b) 69
(c) 87
(d) 78

Answer:

The correct option is (d).

Let the tens and the units digits of the required number be x and y, respectively.
Required number = (10x + y)
According to the question, we have:
x + y = 15 ....(i)
Number obtained on reversing its digits = (10y + x)
∴ (10y + x) = (10x + y) + 9
⇒ 10y + x − 10x − y = 9
⇒ 9y − 9x = 9
⇒ y − x = 1 ....(ii)
On adding (i) and (ii), we get:
2y = 16 ⇒ y = 8
On substituting y = 8 in (i), we get:
x + 8 = 15 ⇒ x = (15 − 8) = 7
Number = (10x + y) = 10 × 7 + 8 = 70 + 8 = 78
Hence, the required number is 78.

Page No 160:

Question 23:

In a given fraction, if 1 is subtracted from the numerator and 2 is added to the denominator, it becomes $\frac{1}{2}$ . If 7 is subtracted from the numerator and 2 is subtracted from the denominator, it becomes $\frac{1}{3}$ . The fraction is
(a) $\frac{13}{24}$

(b) $\frac{15}{26}$

(c) $\frac{16}{27}$

(d) $\frac{16}{21}$

Answer:

Let the fraction be $\frac{x}{y}$
It is given that $\frac{x - 1}{y + 2} = \frac{1}{2}$
$\Rightarrow 2 x - 2 = y + 2 \Rightarrow 2 x - y = 4 . . . . . (i)$
Also, $\frac{x - 7}{y - 2} = \frac{1}{3}$
$\Rightarrow 3 x - 21 = y - 2 \Rightarrow 3 x - y = 19 . . . . . (ii)$
Subtract (ii) from (i), we get
$x = 15$
Put the value of x in equation (i), we get;
$2 (15) - y = 4 30 - y = 4 \Rightarrow y = 30 - 4 = 26$
Therefore, the fraction is $\frac{x}{y} = \frac{15}{26}$
Hence, the correct answer is option (b)

Page No 160:

Question 24:

5 years hence, the age of a man shall be 3 times the age of his son while 5 years earlier the age of the man was 7 times the age of his son. The present age of the man is
(a) 45 years
(b) 50 years
(c) 47 years
(d) 40 years

Answer:

The correct option is (d).
Let the man's present age be x years.
Let his son's present age be y years.
Five years later:
(x + 5) = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x − 3y = 10 ....(i)
Five years ago:
(x − 5) = 7(y − 5)
⇒ x − 5 = 7y − 35
⇒ x − 7y = −30 ....(ii)
On subtracting (i) from (ii), we get:
−4y = −40 ⇒ y = 10
On substituting y = 10 in (i), we get:
x − 3 × 10 = 10 ⇒ x − 30 = 10 ⇒ x = (10 + 30) = 40 years
Hence, the man's present age is 40 years.

Page No 160:

Question 25:

The graphs of the equations $6 x - 2 y + 9 = 0 and 3 x - y + 12 = 0$ are two lines which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer:

The correct option is (b).

The given equations are as follows:
$6 x - 2 y + 9 = 0 and 3 x - y + 12 = 0$
They are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 6, b₁ = −2, c₁ = 9 and a₂ = 3, b₂ = −1 and c₂ = 12
∴ $\frac{a_{1}}{a_{2}} = \frac{6}{3} = \frac{2}{1}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 1} = \frac{2}{1} and \frac{c_{1}}{c_{2}} = \frac{9}{12} = \frac{3}{4}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The given system has no solution.
Hence, the lines are parallel.

Page No 160:

Question 26:

The graphs of the equations $2 x + 3 y - 2 = 0 and x - 2 y - 8 = 0$ are two which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer:

The correct option is (c).

The given equations are as follows:
$2 x + 3 y - 2 = 0 and x - 2 y - 8 = 0$
They are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = 3, c₁ = −2 and a₂ = 1, b₂ = −2 and c₂ = −8
∴ $\frac{a_{1}}{a_{2}} = \frac{2}{1}, \frac{b_{1}}{b_{2}} = \frac{3}{- 2} and \frac{c_{1}}{c_{2}} = \frac{- 2}{- 8} = \frac{1}{4}$
∴ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
The given system has a unique solution.
Hence, the lines intersect exactly at one point.

Page No 160:

Question 27:

The graphs of the equations $5 x - 15 y = 8 and 3 x - 9 y = \frac{24}{5}$ are two which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer:

The correct option is (a).

The given system of equations can be written as follows:
$5 x - 15 y - 8 = 0 and 3 x - 9 y - \frac{24}{5} = 0$
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 5, b₁ = −15, c₁ = −8 and a₂ = 3, b₂ = −9 and c₂ = $- \frac{24}{5}$
∴ $\frac{a_{1}}{a_{2}} = \frac{5}{3}, \frac{b_{1}}{b_{2}} = \frac{- 15}{- 9} = \frac{5}{3} and \frac{c_{1}}{c_{2}} = - 8 \times \frac{5}{- 24} = \frac{5}{3}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
The given system of equations will have an infinite number of solutions.
Hence, the lines are coincident.

Page No 163:

Question 1:

The graphic representation of the equations $x + 2 y = 3 and 2 x + 4 y + 7 = 0$ gives a pair of
(a) parallel lines
(b) intersecting lines
(c) coincident lines
(d) none of these

Answer:

The correct option is (a).

The given system of equations can be written as follows:
x + 2y − 3 = 0 and 2x + 4y + 7 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = 2, c₁ = −3 and a₂ = 2, b₂ = 4 and c₂ = 7
∴ $\frac{a_{1}}{a_{2}} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{2}{4} = \frac{1}{2} and \frac{c_{1}}{c_{2}} = \frac{- 3}{7}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
So, the given system has no solution.
Hence, the lines are parallel.

Page No 163:

Question 2:

If $2 x - 3 y = 7 and (a + b) x - (a + b - 3) y = 4 a + b$ have an infinite number of solutions, then
(a) a = 5, b = 1
(b) a = −5, b = 1
(c) a = 5, b = −1
(d) a = −5, b = −1

Answer:

The correct option is (d).

The given system of equations can be written as follows:
2x − 3y − 7 = 0 and (a + b)x − (a + b − 3)y − (4a + b) = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = −3, c₁ = −7 and a₂ = (a + b), b₂ = −(a + b − 3) and c₂ = −(4a + b)
∴ $\frac{a_{1}}{a_{2}} = \frac{2}{(a + b)}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- (a + b - 3)} = \frac{3}{(a + b - 3)} and \frac{c_{1}}{c_{2}} = \frac{- 7}{- (4 a + b)} = \frac{7}{(4 a + b)}$
For an infinite number of solutions, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
∴ $\frac{2}{(a + b)} = \frac{3}{(a + b - 3)} = \frac{7}{(4 a + b)}$
Now, we have:
$\frac{2}{(a + b)} = \frac{3}{(a + b - 3)} \Rightarrow 2 a + 2 b - 6 = 3 a + 3 b$
⇒ a + b + 6 = 0                    ...(i)
Again, we have:
$\frac{3}{(a + b - 3)} = \frac{7}{(4 a + b)} \Rightarrow 12 a + 3 b = 7 a + 7 b - 21$
⇒ 5a − 4b + 21 = 0                            ...(ii)

On multiplying (i) by 4, we get:
4a + 4b + 24 = 0                              ...(iii)
On adding (ii) and (iii), we get:
9a = −45 ⇒ a = −5
On substituting a = −5 in (i), we get:
−5 + b + 6 = 0 ⇒ b = −1
∴ a = −5 and b = −1

Page No 163:

Question 3:

The pair of equations $2 x + y = 5, 3 x + 2 y = 8$ has
(a) a unique solution
(b) two solutions
(c) no solution
(d) infinitely many solutions

Answer:

The correct option is (a).

The given system of equations can be written as follows:
2x + y − 5 = 0 and 3x + 2y − 8 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 2, b₁ = 1, c₁ = −5 and a₂ = 3, b₂ = 2 and c₂ = −8
∴ $\frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{2} and \frac{c_{1}}{c_{2}} = \frac{- 5}{- 8} = \frac{5}{8}$
∴ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
The given system has a unique solution.
Hence, the lines intersect at one point.

Page No 163:

Question 4:

If x = −y and y > 0, which of the following is wrong?
(a) $x^{2} y > 0$
(b) $x + y = 0$
(c) $x y < 0$
(d) $\frac{1}{x} - \frac{1}{y} = 0$

Answer:

The correct option is (d).

Given:
x = −y and y > 0
Now, we have:
(i) x²y
On substituting x = −y, we get:
(−y)²y = y³ > 0 (∵ y > 0)
This is true.

(ii) x + y
On substituting x = −y, we get:
(−y) + y = 0
This is also true.

(iii) xy
On substituting x = −y, we get:
(−y) y = −y² < 0 (∵ y > 0)
This is again true.

(iv) $\frac{1}{x} - \frac{1}{y} = 0$
$\Rightarrow \frac{y - x}{x y} = 0$
On substituting x = −y, we get:
$\frac{y - (- y)}{(- y) y} = 0 \Rightarrow \frac{2 y}{- y^{2}} = 0 \Rightarrow 2 y = 0 \Rightarrow y = 0$
Hence, from the above equation, we get y = 0, which is wrong.

Page No 163:

Question 5:

Show that the system of equations $- x + 2 y + 2 = 0 and \frac{1}{2} x - \frac{1}{4} y - 1 = 0$ has a unique solution.

Answer:

The given system of equations:
$- x + 2 y + 2 = 0 and \frac{1}{2} x - \frac{1}{4} y - 1 = 0$
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = −1, b₁ = 2, c₁ = 2 and a₂ = $\frac{1}{2}$ , b₂ = $- \frac{1}{4}$ and c₂ = −1
∴ $\frac{a_{1}}{a_{2}} = \frac{- 1}{(\frac{1}{2})} = - 2, \frac{b_{1}}{b_{2}} = \frac{2}{(- \frac{1}{4})} = - 8 and \frac{c_{1}}{c_{2}} = \frac{2}{- 1} = - 2$
∴ $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
The given system has a unique solution.
Hence, the lines intersect at one point.

Page No 163:

Question 6:

For what values of k is the system of equations $k x + 3 y = k - 2, 12 x + k y = k$ inconsistent?

Answer:

The given system of equations can be written as follows:
kx + 3y − (k − 2) = 0 and 12x + ky − k = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁ = 3, c₁ = − (k − 2) and a₂ = 12, b₂ = k and c₂ = − k
∴ $\frac{a_{1}}{a_{2}} = \frac{k}{12}, \frac{b_{1}}{b_{2}} = \frac{3}{k} and \frac{c_{1}}{c_{2}} = \frac{- (k - 2)}{- k} = \frac{(k - 2)}{k}$
For inconsistency, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
⇒ $\frac{k}{12} = \frac{3}{k} \neq \frac{(k - 2)}{k} \Rightarrow k^{2} = (3 \times 12) = 36$
⇒ $k = \sqrt{36} = \pm 6$
Hence, the pair of equations is inconsistent if $k = \pm 6$ .

Page No 163:

Question 7:

Show that the equations $9 x - 10 y = 21, \frac{3 x}{2} - \frac{5 y}{3} = \frac{7}{2}$ have infinitely many solutions.

Answer:

The given system of equations can be written as follows:
$9 x - 10 y - 21 = 0 and \frac{3 x}{2} - \frac{5 y}{3} - \frac{7}{2} = 0$
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 9, b₁ = −10, c₁ = −21 and a₂ = $\frac{3}{2}$ , b₂ = $\frac{- 5}{3}$ and c₂ = $\frac{- 7}{2}$
∴ $\frac{a_{1}}{a_{2}} = \frac{9}{\frac{3}{2}} = 6, \frac{b_{1}}{b_{2}} = \frac{- 10}{(\frac{- 5}{3})} = 6 and \frac{c_{1}}{c_{2}} = - 21 \times \frac{2}{- 7} = 6$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
This shows that the given system equations has an infinite number of solutions.

Page No 163:

Question 8:

Solve the system of equations: $x - 2 y = 0, 3 x + 4 y = 20$

Answer:

The given equations are as follows:
x − 2y = 0                     ....(i)
3x + 4y = 20                 ....(ii)
On multiplying (i) by 2, we get:
2x − 4y = 0                        ....(iii)
On adding (ii) and (iii), we get:
5x = 20 ⇒ x = 4
On substituting x = 4 in (i), we get:
4 − 2y = 0 ⇒ 4 = 2y ⇒ y = 2
Hence, the required solution is x = 4 and y = 2.

Page No 163:

Question 9:

Show that the paths represented by the equations $x - 3 y = 2 and - 2 x + 6 y = 5$ are parallel.

Answer:

The given system of equations can be written as follows:
x − 3y − 2 = 0 and −2x + 6y − 5= 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = −3, c₁ = −2 and a₂ = −2, b₂ = 6 and c₂ = −5
∴ $\frac{a_{1}}{a_{2}} = \frac{1}{- 2} = \frac{- 1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 3}{6} = \frac{- 1}{2} and \frac{c_{1}}{c_{2}} = \frac{- 2}{- 5} = \frac{2}{5}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Thus, the given system of equations has no solution.
Hence, the paths represented by the equations are parallel.

Page No 163:

Question 10:

The difference between two numbers is 26 and one number is three times the other. Find the numbers.

Answer:

Let the larger number be x and the smaller number be y.
Then, we have:
x − y = 26 ...(i)
x = 3y ...(ii)
On substituting x = 3y in (i), we get:
3y − y = 26 ⇒ 2y = 26 ⇒ y = 13
On substituting y = 13 in (i), we get:
x − 13 = 26 ⇒ x = 26 + 13 = 39
Hence, the required numbers are 39 and 13.

Page No 163:

Question 11:

Solve: $23 x + 29 y = 98, 29 x + 23 y = 110$

Answer:

The given equations are as follows:
23x + 29y = 98                              ....(i)
29x + 23y = 110                          ....(ii)
On adding (i) and (ii), we get:
52x + 52y = 208
⇒ x + y = 4                           ....(iii)
On subtracting (i) from (ii), we get:
6x − 6y = 12
⇒ x − y = 2                       ....(iv)
On adding (iii) and (iv), we get:
2x = 6 ⇒ x = 3
On substituting x = 3 in (iii), we get:
3 + y = 4
⇒ y = 4 − 3 = 1
Hence, the required solution is x = 3 and y = 1.

Page No 163:

Question 12:

Solve: $6 x + 3 y = 7 x y and 3 x + 9 y = 11 x y$

Answer:

The given equations are as follows:
6x + 3y = 7xy                   ....(i)
3x + 9y = 11xy                     ....(ii)

For equation (i), we have:

$\frac{6 x + 3 y}{x y} = 7$
$\Rightarrow \frac{6 x}{x y} + \frac{3 y}{x y} = 7 \Rightarrow \frac{6}{y} + \frac{3}{x} = 7$       ....(iii)

For equation (ii), we have:

$\frac{3 x + 9 y}{x y} = 11$
$\Rightarrow \frac{3 x}{x y} + \frac{9 y}{x y} = 11 \Rightarrow \frac{3}{y} + \frac{9}{x} = 11$       ....(iv)
On substituting $\frac{1}{y} = v$ and $\frac{1}{x} = u$ in (iii) and (iv), we get:
6v + 3u = 7                              ....(v)
3v + 9u = 11                            ....(vi)
On multiplying (v) by 3, we get:
18v + 9u = 21                          ....(vii)
On subtracting (vi) from (vii), we get:
15v = 10 ⇒ v = $\frac{10}{15} = \frac{2}{3}$
⇒ $\frac{1}{y} = \frac{2}{3} \Rightarrow y = \frac{3}{2}$
On substituting $y = \frac{3}{2}$ in (iii), we get:
$\frac{6}{(\frac{3}{2})} + \frac{3}{x} = 7$
$\Rightarrow 4 + \frac{3}{x} = 7 \Rightarrow \frac{3}{x} = 3 \Rightarrow 3 x = 3$
$\Rightarrow x = 1$
Hence, the required solution is x = 1 and $y = \frac{3}{2}$ .

Page No 163:

Question 13:

Find the value of k for which the system of equations $3 x + y = 1 and k x + 2 y = 5$ has (i) a unique solution, (ii) no solution.

Answer:

The given system of equations:
3x + y = 1
⇒ 3x + y − 1= 0 ....(i)
kx + 2y = 5
⇒ kx + 2y − 5 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁= 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁= 1, c₁ = −1 and a₂ = k, b₂= 2, c₂ = −5

(i) For a unique solution, we must have:
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ i.e. $\frac{3}{k} \neq \frac{1}{2} \Rightarrow k \neq 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.

(ii) In order that the given equations have no solution, we must have:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
$\frac{3}{k} = \frac{1}{2} \neq \frac{- 1}{- 5}$
$\Rightarrow \frac{3}{k} = \frac{1}{2} and \frac{3}{k} \neq \frac{- 1}{- 5}$
$\Rightarrow k = 6, k \neq 15$
Thus, for k = 6, the given system of equations will have no solution.

Page No 163:

Question 14:

In $∆ A B C, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B),$ find the measure of each one of $∠ A, ∠ B and ∠ C$ .

Answer:

Let $∠ A = x ° and ∠ B = y °$
Then, $∠ C = 3 ∠ B = (3 y) °$
Now, we have:
$∠ A + ∠ B + ∠ C = 180 °$
⇒ x + y + 3y = 180
⇒ x + 4y = 180              ....(i)
Also, $∠ C = 2 (∠ A + ∠ B)$
⇒ 3y = 2(x + y)
⇒ 2x − y = 0                ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 0                    ....(iii)
On adding (i) and (iii), we get:
9x = 180 ⇒ x = 20
On substituting x = 20 in (i), we get:
20 + 4y = 180 ⇒ 4y = (180 − 20) = 160 ⇒ y = 40
∴ x = 20 and y = 40
∴ $∠ A = 20 °, ∠ B = 40 °, ∠ C = (3 \times 40) ° = 120 °$

Page No 164:

Question 15:

5 pencils and 7 pens together cost Rs 195 while 7 pencils and 5 pens together cost Rs 153. Find the cost of each one of the pencil and the pen.

Answer:

Let the cost of each pencil be Rs. x and that of each pen be Rs. y.
Then, we have:
5x + 7y = 195                       ....(i)
7x + 5y = 153                   ....(ii)
Adding (i) and (ii), we get:
12x + 12y = 348
⇒ 12(x + y) = 348
⇒ x + y = 29                             ....(iii)
Subtracting (i) from (ii), we get:
2x − 2y = −42
⇒ 2(x − y) = −42
⇒ x − y = −21                          ....(iv)
On adding (iii) and (iv), we get:
2x = 8 ⇒ x = 4
On substituting x = 4 in (iii), we get:
4 + y = 29 ⇒ y = (29 − 4) = 25
Hence, the cost of each pencil is Rs. 4 and the cost of each pen is Rs. 25.

Page No 164:

Question 16:

Solve the following system of equations graphically:
$2 x - 3 y = 1, 4 x - 3 y + 1 = 0$

Answer:

On a graph paper, draw a horizontal line X'OX and a vertical line YOY' as the x-axis and the y-axis, respectively.

Graph of 2x − 3y = 1
2x − 3y = 1
⇒ 3y = (2x − 1)
∴ $y = \frac{2 x - 1}{3}$ ...(i)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 1
Putting x = 5, we get:
y = 3
Thus, we have the following table for the equation 2x − 3y = 1.

x	−1	2	5
y	−1	1	3

Now, plots the points A(−1, −1), B(2, 1) and C(5, 3) on the graph paper.
Join AB and BC to get the graph line AC. Extend it on both the sides.
Thus, the line AC is the graph of 2x − 3y = 1.

Graph of 4x − 3y + 1 = 0
4x − 3y + 1 = 0
⇒ 3y = ( 4x + 1)
∴

y = \frac{4 x + 1}{3}

...(ii)
Putting x = −1, we get:
y = −1
Putting x = 2, we get:
y = 3
Putting x = 5, we get:
y = 7
Thus, we have the following table for the equation 4x − 3y + 1 = 0.

x	−1	2	5
y	−1	3	7

Now, plots the points P(2, 3) and Q(5, 7). The point A(−1, −1) has already been plotted. Join PA and QP to get the graph line AQ. Extend it on both the sides.
Thus, the line AQ is the graph of the equation 4x − 3y + 1 = 0.

The two lines intersect at A(−1, −1).
Thus, x = −1 and y = −1 is the solution of the given system of equations.

Page No 164:

Question 17:

Find the angles of a cyclic quadrilateral ABCD in which $∠ A = (4 x + 20) °, ∠ B = (3 x - 5) °, ∠ C = (4 y) ° and ∠ D = (7 y + 5) ° .$

Answer:

Given:
In a cyclic quadrilateral ABCD, we have:
$∠ A = (4 x + 20) °$
$∠ B = (3 x - 5) °$
$∠ C = (4 y) °$
$∠ D = (7 y + 5) °$
$∠ A + ∠ C = 180 °$ and $∠ B + ∠ D = 180 °$       [Since ABCD is a cyclic quadrilateral]
Now, $∠ A + ∠ C = (4 x + 20) ° + (4 y) ° = 180 °$
⇒ 4x + 4y + 20 = 180
⇒ 4x + 4y = 180 − 20 = 160
⇒ x + y = 40                       ....(i)
Also, $∠ B + ∠ D = (3 x - 5) ° + (7 y + 5) ° = 180 °$
⇒ 3x + 7y = 180                 ....(ii)
On multiplying (i) by 3, we get:
3x + 3y = 120                     ....(iii)
On subtracting (iii) from (ii), we get:
4y = 60 ⇒ y = 15
On substituting y = 15 in (1), we get:
x + 15 = 40 ⇒ x = (40 − 15) = 25
Therefore, we have:
$∠ A = (4 x + 20) ° = (4 \times 25 + 20) ° = 120 °$
$∠ B = (3 x - 5) ° = (3 \times 25 - 5) ° = 70 °$
$∠ C = (4 y) ° = (4 \times 15) ° = 60 °$
$∠ D = (7 y + 5) ° = (7 \times 15 + 5) ° = (105 + 5) ° = 110 °$

Page No 164:

Question 18:

Solve for x and y: $\frac{35}{x + y} + \frac{14}{x - y} = 19, \frac{14}{x + y} + \frac{35}{x - y} = 37$

Answer:

We have:
$\frac{35}{x + y} + \frac{14}{x - y} = 19 and \frac{14}{x + y} + \frac{35}{x - y} = 37$
Taking $\frac{1}{x + y} = u$ and $\frac{1}{x - y} = v$ :
35u + 14v − 19 = 0                    ....(i)
14u + 35v − 37 = 0                    ....(ii)
Here, a₁= 35, b₁ = 14, c₁ = −19, a₂ = 14, b₂ = 35, c₂ = −37
By cross multiplication, we have:

∴ $\frac{u}{[14 \times (- 37) - 35 \times (- 19)]} = \frac{v}{[(- 19) \times 14 - (- 37) \times (35)]} = \frac{1}{[35 \times 35 - 14 \times 14]}$
⇒ $\frac{u}{- 518 + 665} = \frac{v}{- 266 + 1295} = \frac{1}{1225 - 196}$
⇒ $\frac{u}{147} = \frac{v}{1029} = \frac{1}{1029}$
⇒ $u = \frac{147}{1029} = \frac{1}{7}, v = \frac{1029}{1029} = 1$
⇒ $\frac{1}{x + y} = \frac{1}{7}, \frac{1}{x - y} = 1$
∴ (x + y) = 7                          ....(iii)
And, (x − y) = 1 ....(iv)

Again, the equations (iii) and (iv) can be written as follows:
x + y − 7 = 0                               ....(v)
x − y − 1 = 0 ....(vi)
Here, a₁= 1, b₁ = 1, c₁ = −7 , a₂ = 1 , b₂ = −1 , c₂ = −1
By cross multiplication, we have:

⇒ $\frac{x}{[1 \times (- 1) - (- 1) \times (- 7)]} = \frac{y}{[(- 7) \times 1 - (- 1) \times 1]} = \frac{1}{[1 \times (- 1) - 1 \times 1]}$

⇒ $\frac{x}{- 1 - 7} = \frac{y}{- 7 + 1} = \frac{1}{- 1 - 1}$
⇒ $\frac{x}{- 8} = \frac{y}{- 6} = \frac{1}{- 2}$
⇒ $x = \frac{- 8}{- 2} = 4, y = \frac{- 6}{- 2} = 3$
Hence, x = 4 and y = 3 is the required solution.

Page No 164:

Question 19:

If 1 is added to both of the numerator and denominator of a fraction, it becomes $\frac{4}{5}$ . If however, 5 is subtracted from both numerator and denominator, the fraction $\frac{1}{2}$ . Find the fraction.

Answer:

Let the required fraction be $\frac{x}{y}$ .
Then, we have:
$\frac{x + 1}{y + 1} = \frac{4}{5}$
⇒ 5(x + 1) = 4(y + 1)
⇒ 5x + 5 = 4y + 4
⇒ 5x − 4y = −1                       ....(i)

Again, we have:
$\frac{x - 5}{y - 5} = \frac{1}{2}$
⇒ 2(x − 5) = 1(y − 5)
⇒ 2x − 10 = y − 5
⇒ 2x − y = 5                         ....(ii)
On multiplying (ii) by 4, we get:
8x − 4y = 20                             ....(iii)
On subtracting (i) from (iii), we get:
3x = (20 − (−1)) = 20 + 1 = 21
⇒ 3x = 21
⇒ x = 7
On substituting x = 7 in (i), we get:
5 × 7 − 4y = −1
⇒ 35 − 4y = −1
⇒ 4y = 36
⇒ y = 9
∴ x = 7 and y = 9
Hence, the required fraction is $\frac{7}{9}$ .

Page No 164:

Question 20:

Solve: $\frac{a x}{b} - \frac{b y}{a} = a + b, a x - b y = 2 a b$

Answer:

The given equations may be written as follows:
$\frac{a x}{b} - \frac{b y}{a} - (a + b) = 0$ ....(i)
$a x - b y - 2 a b = 0$ ....(ii)
Here, a₁= $\frac{a}{b}$ , b₁ = $\frac{- b}{a}$ , c₁ = −(a + b), a₂ = a, b₂ = −b, c₂ = −2ab
By cross multiplication, we have:

∴ $\frac{x}{(- \frac{b}{a}) \times (- 2 a b) - (- b) \times (- (a + b))} = \frac{y}{- (a + b) \times a - (- 2 a b) \times \frac{a}{b}} = \frac{1}{\frac{a}{b} \times (- b) - a \times (- \frac{b}{a})}$
⇒ $\frac{x}{2 b^{2} - b (a + b)} = \frac{y}{- a (a + b) + 2 a^{2}} = \frac{1}{- a + b}$
⇒ $\frac{x}{2 b^{2} - a b - b^{2}} = \frac{y}{- a^{2} - a b + 2 a^{2}} = \frac{1}{- a + b}$
⇒ $\frac{x}{b^{2} - a b} = \frac{y}{a^{2} - a b} = \frac{1}{- (a - b)}$
⇒ $\frac{x}{- b (a - b)} = \frac{y}{a (a - b)} = \frac{1}{- (a - b)}$
⇒ $x = \frac{- b (a - b)}{- (a - b)} = b, y = \frac{a (a - b)}{- (a - b)} = - a$
Hence, x = b and y = −a is the required solution.

View NCERT Solutions for all chapters of Class 10