RD Sharma 2022 Solutions for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables

Answer:

Let no. of ride is   and no. of Hoopla is .He paid Rs 20 for ride and for Hoopla.

The cost of ride is Rs and cost of Hoopla is Rs.then

The number of Hoopla is the half number of ride, then

Hence algebraic equations are and

Now, we draw the graph for algebraic equations.

Answer:

Let age of Aftab is  years and age of his daughter is years. Years ago his age wastimes older as her daughter was. Then

Three years from now, he will be three times older as his daughter will be, then

Hence the algebraic representation are and

Answer:

The given equation are   and.

In order to represent the above pair of linear equation graphically, we need

Two points on the line representing each equation. That is, we find two solutions

of each equation as given below:

We have,

Putting we get

Putting we get

Thus, two solution of equation are

We have

Putting we get

Putting we get

Thus, two solution of equation are

		8
	6

Now we plot the pointand and draw a line passing through

These two points to get the graph o the line represented by equation

We also plot the points and and draw a line passing through

These two points to get the graph O the line represented by equation

We observe that the line parallel and they do not intersect anywhere.

Answer:

Gloria is walking the path joining and

Suresh is walking the path joining and

	0	4
	5	0

The graphical representations are

Answer:

(i) Given equation are: 5x + 4y + 8 = 0 

7x + 6y − 9 = 0

We have And

Thus the pair of linear equation is intersecting.

(ii) Given equation are: 9x + 3y + 12 = 0

18x + 6y + 24 = 0

We have

Thus the pair of linear is coincident lines.

(iii) Given equation are:

We have

Thus the pair of line is parallel lines.

Answer:

(i) Given the linear equation are:

We know that intersecting condition:

Where

Hence the equation of other line is

(ii) We know that parallel line condition is:

Where

Hence the equation is

(iii) We know that coincident line condition is:

Where

Hence the equation is

Answer:

Let the cost of 1 kg of apples be Rs x.

And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

For ,

The solution table is

For 4x + 2y = 300,

The solution table is

The graphical representation is as follows.

Answer:

The given equations are:

x – y + 1 = 0    ...(i)
3x + 2y – 12 = 0     ...(ii)

Solving equation (i), we get

$$\begin{gathered} x-y+1=0 \\ \Rightarrow y=x+1 \end{gathered}$$

When x = 0, y = 1
When x = 1, y = 2
 

Answer:

The given equations are:

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equationwe get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at points P.

Hence, and is the solution.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table. 

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Thus the graph of the two equations are coincide 

Consequently, every solution of one equation is a solution of the other.

Hence the equations have infinitely many solutions.

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here we see that the two lines are parallel 

Hence the given system of equations has no solution.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points_.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Here, the two lines are parallel.

Hence the given system of equations is inconsistent.

Answer:

(i) Draw the 3 lines as given by equations 

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

Clearly from graph points of intersection three lines are

(−4,2) , (1,3), (2,5) 

(ii) Draw the 3 lines as given by equations 

By taking x=1 = 1 cm on x−axis

And y =1=1cm on y−axis

From graph point of intersection are (0,0) (10/3,0) (5/3,5/3)

Answer:

The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

	0	5/4
	–5/2	0

Draw the graph by plotting the two points from table.

It has unique solution.

Hence the system of equations is consistent

Answer:

(i) The given equations are

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Graph of the equation….

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at point .

Hence the equations have unique solution.

(ii) The equations of graphs is

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

The graph of (i) can be obtained by plotting the two points .

Graph of the equation

Putting in equation we get.

Putting in equation we get.

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines are coincident.

Hence the equations have infinitely much solution.

Hence the system is consistent

Answer:

​(i) The given equations are 

_{The two points satisfying (i) can be listed in a table as,}

	0	8
	$\frac{4}{5}$	4

_{The two points satisfying (ii) can be listed in a table as,}

	4	2	0
	0	4	8

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 3, y = 2.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

(ii) The given equations are 

_{The two points satisfying (i) can be listed in a table as,}

	4	0
	0	6

_{The two points satisfying (ii) can be listed in a table as,}

	0	3
	−2	5.5

Now, graph of equations (i) and (ii) can be drawn as,

It is seen that the solution of the given system of equations is given by x = 2, y = 3.

Also, it is observed that the lines (i) and (ii) meet the y-axis at the points respectively.

Answer:

​(i) The given equations are:

Putting  in equation we get:

Putting  in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting  in equation we get:

Putting  in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points  from table.

The two lines intersect at. The region enclosed by the lines represented by the given equations and x-axis are shown in the above figure.

Hence,  and  is the solution.

(ii) The given equations are:

Putting  in equation we get:

Putting  in equationwe get:

Use the following table to draw the graph.

	0	$\frac{4}{3}$
	2	0

The graph of (i) can be obtained by plotting the two points $A\left(0, 2\right), B\left(\frac{4}{3}, 0\right)$.

Putting  in equation we get:

Putting  in equationwe get:

Use the following table to draw the graph.

		$\frac{7}{2}$
	−$\frac{7}{3}$

Draw the graph by plotting the two points  from table. 

The two lines intersect at. The area enclosed by the lines represented by the given equations and the coordinates x-axis and shaded the area in graph.

Hence,  and is the solution.

Answer:

The given equations are

Putting in equation we get:

Putting in equation we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Draw the graph by plotting the two points from table.

The intersection point is P(3, 2)

Three points of the triangle are. 

Hence the value of and

Answer:

The given equations are

Putting in equation we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table.

The two lines intersect at.

Now, Required area = Area of shaded region

Required area = Area of PBD

Required area =

Required area =

Required area =

Hence the area =

Answer:

The given equations are: 

Putting in equation (i) we get:

Putting in equation (i) we get:

Use the following table to draw the graph.

Draw the graph by plotting the two points from table

Putting in equation we get:

Putting in equationwe get:

Use the following table to draw the graph.

	–3

Draw the graph by plotting the two points from table.

Hence the vertices of the required triangle are.

Now, 

Required area = Area of PCA

Required area =

Required area =

Hence the required area is

Answer:

​(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y 

For x − y = 4,

x = 4 + y

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be ₹ x and the cost of 1 pen be ₹ y.

According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

7x + 5y = 46

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are ₹ 3 and ₹ 5 respectively.

Answer:

The given equations are 

_{The two points satisfying (i) can be listed in a table as,}

	−3	6
	0	6

The two points satisfying (ii) can be listed in a table as,

	0	9
	6	0

The two points satisfying (iii) can be listed in a table as,

	−1	8
	2	2

Now, graph of equations (i), (ii) and (iii) can be drawn as,

It is seen that the coordinates of the vertices of the obtained triangle are 

_{∴Area of ΔABC =}

Answer:

​Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are :

y = 2x − 2 … (i)

y = 4x − 4 … (ii)

The graphs of the equations (i) and (ii) can be drawn by finding two solutions for each of the equations. They are given in the following table.

Hence, the graphic representation is as follows.

The two lines intersect at the point (1, 0).

So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

Answer:

(i) For intersecting lines,

Equation of another intersecting line to the given line is−

Since, condition for intersecting lines and unique solution is−

(ii) For parallel lines,

Equation of another parallel line to the given line is−

Since, condition for parallel lines and no solution is−

(iii) For co−incident lines,

Equation of another coincident line to the given line is−

Since, condition for coincident lines and infinite solution is−

Answer:

The given linear equations are:
$$\begin{gathered} 2x+y=6 .....\left(\mathrm{i}\right) \\ 2x-y+2=0 .....\left(\mathrm{ii}\right) \end{gathered}$$
For (i), we have

Answer:

The given lines are:
$$\begin{gathered} y=x .....\left(\mathrm{i}\right) \\ 3y=x .....\left(\mathrm{ii}\right) \\ x+y=8 .....\left(\mathrm{iii}\right) \end{gathered}$$
From (i), we have

Answer:

The given equations are 
x = 3                              .....(i)
x = 5                              .....(ii)
2x − y − 4 = 0                .....(iii)
For (iii), we have

Answer:

The given lines are 
$$\begin{gathered} x=-2 .....\left(\mathrm{i}\right) \\ y=3 .....\left(\mathrm{ii}\right) \end{gathered}$$
The graph thus obtained will be as follows:

The figure thus obtained is a rectangle ABCD. The vertices of the rectangle are A(−2, 3), B(0, 3), C(0, 0) and D(−2, 0).
Length of the rectangle = AD = BC = 3 units
Breadth of the rectangle = CD = AB =  2 units
∴ Area of rectangle ABCD
$$\begin{gathered} =3\times 2 \\ =6 \mathrm{square} \mathrm{units} \end{gathered}$$

Answer:

ABCD is the given rectangle. So, AB = CD and AD = BC.
Thus,
$$\begin{gathered} x+3y=13 .....\left(\mathrm{i}\right) \\ 3x+y=7 .....\left(\mathrm{ii}\right) \end{gathered}$$
Adding (i) and (ii), we get
4x + 4y = 20    
$\Rightarrow x+y=5 .....\left(\mathrm{iii}\right)$
Subtracting (i) from (ii), we get
2x − 2y = −6
$\Rightarrow x-y=-3 .....\left(\mathrm{iv}\right)$
Adding (iii) and (iv), we get
2x = 2
$\Rightarrow x=1$
Putting x = 1 in (iii), we get
1 + y = 5
⇒ y = 4
Thus, x = 1 and y = 4.

Answer:

The given equations are:

Multiply equation by 2 and equation by 15, and add both equations we get

Put the value of in equationwe get

Hence the value of and

Answer:

The given equations are:

…

Subtract (ii) from (i) we get

Put the value of in equation we get

Hence the value of and .

Answer:

The given equations are:

…

…

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

Answer:

The given equations are:

…

…

Multiply equation by and equation by and add both equations we get

Put the value of in equation we get

Hence the value of and .

Answer:

The given equations are:

$$\begin{gathered} 3x - \frac{y + 7}{11} + 2 = 10 \\ \Rightarrow 3x - \frac{\left(y + 7\right)}{11} = 8 \\ \Rightarrow \frac{33x - y - 7 }{11} = 8 \\ \Rightarrow 33x - y - 7 = 88 \\ \Rightarrow 33x - y = 95 ........\left(1\right) \end{gathered}$$ 

$$\begin{gathered} 2y + \frac{x + 11}{7} = 10 \\ \Rightarrow \frac{14y + x + 11}{7} = 10 \\ \Rightarrow 14y + x + 11 = 70 \\ \Rightarrow 14y + x = 59 \\ \Rightarrow x + 14y = 59 .........\left(2\right) \end{gathered}$$

Multiply equation (1) by , we get

$462x - 14y = 1330 ......\left(3\right)$

Hence the value of x and y are and

Answer:

The given equations are:

…

…

Multiply equation by and equation by, add both equations, we get

Put the value of in equation, we get

Hence the value of and .

Answer:

The given equations are:

…

…

Multiply equation by and equation by 3 and add both equations we get

Put the value of in equation, we get

Hence the value of and .

Answer:

The given equations are:

…

…

Multiply equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

…

…

Multiply equation by and by and subtract equation (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

…

…

Multiply equation by and equation by and subtract equation (ii) from (i), we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

Add both equations, we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

…

…

Add both equations we get

Put the value of in equation, we get

Hence the value of and .

Answer:

The given equations are:

Let and then equations are

…

…

Multiply equation by and equation by add both equations, we get

Put the value of in equation, we get

Then 

…

…

Add both equations, we get

Put the value of in equation we get

Hence the value of and

Answer:

The given equations are:

Let and then equations are 

…

…

Multiply equation by and add both equations, we get

Put the value of in equation, we get 

Then 

Hence the value of and .

Answer:

The given equations are:

Let and then equations are

…

…

Multiply equation by and equation by and add both equations, we get

Put the value of in equation, we get

Then

Add both equations, we get

Put the value of in equation we get

Hence the value of and .

Answer:

The given equations are:

Let and then equations are

…

…

Multiply equation by and add both equations, we get

Put the value of in equation, we get

Then 

…

…

Add both equations, we get

Put the value of in equation we get

Hence the value of and

Answer:

The given equations are:

…

…

Multiply equation by and equation by, add both equations we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

…

…

Multiply equation by and equation by and add both equations we get

Put the value of in equation, we get

Hence the value of and

Answer:

The given equations are:

…

…

Multiply equation by and equation by, and subtract (ii) from (i) we get

Put the value of in equation, we get

Hence the value of and

Answer:

21x + 47y = 110            .....(i)
47x + 21y = 162            .....(ii)
Adding(i) and (ii), we get
68x + 68y = 272
$\Rightarrow x+y=4$                    .....(iii)
Subtracting (i) from (ii), we get
$$\begin{gathered} 26x-26y=52 \\ \Rightarrow x-y=2 .....\left(\mathrm{iv}\right) \end{gathered}$$               
Adding (iii) and (iv), we get
$$\begin{gathered} 2x=6 \\ \Rightarrow x=3 \end{gathered}$$
Putting x = 3 in (iv), we get
3 − y = 2
⇒ y = 1




 

Answer:

Since (x + 1) is a factor of $2x^3+a{x}^2+2bx+1$, so
$$\begin{gathered} 2{\left(-1\right)}^3+a{\left(-1\right)}^2+2b\left(-1\right)+1=0 \\ \Rightarrow -2+a-2b+1=0 \\ \Rightarrow a-2b-1=0 \\ \Rightarrow a-2b=1 .....\left(\mathrm{i}\right) \end{gathered}$$
Also, we are given
2a − 3b = 4          .....(ii)
From (i) and (ii) we get
$$\begin{gathered} a=1+2b .....\left(\mathrm{iii}\right) \\ \mathrm{Substituting} \mathrm{the} \mathrm{value} \mathrm{of} a \mathrm{in} \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ 2\left(1+2b\right)-3b=4 \\ \Rightarrow 2+4b-3b=4 \\ \Rightarrow b=2 \end{gathered}$$
Putting b = 2 in (iii), we get
a = 1 + 2 × 2 = 5
Thus, the value of a = 5 and b = 2.

Answer:

The given equations are 
$$\begin{gathered} \frac{x}{10}+\frac{y}{5}-1=0 \\ \frac{{\displaystyle x}}{{\displaystyle 8}}+\frac{{\displaystyle y}}{{\displaystyle 6}}=15 \end{gathered}$$
$$\begin{gathered} \frac{x}{10}+\frac{y}{5}=1 \\ \Rightarrow x+2y=10 .....\left(\mathrm{i}\right) \\ \frac{x}{8}+\frac{y}{6}=15 \\ \Rightarrow 3x+4y=360 .....\left(\mathrm{ii}\right) \end{gathered}$$
Multiplying (i) by 2, we get
$2x+4y=20 .....\left(\mathrm{iii}\right)$
Subtracting (ii) from (iii), we get
x = 340
Putting x = 340 in (i), we get
340 + 2y = 10
⇒ 2y = 10 − 340 = −330
⇒ y = −165
Now, in order to find the value of $\lambda$, we simply put the value of x and y in the equation $y=\lambda x+5$.
$$\begin{gathered} \therefore -165=\lambda \left(340\right)+5 \\ \Rightarrow \lambda =-\frac{170}{340} \\ \Rightarrow \lambda =-\frac{1}{2} \end{gathered}$$
Thus, the value of $\lambda =-\frac{1}{2}$.

Answer:

The given equations are
$$\begin{gathered} x+y=2 .....\left(\mathrm{i}\right) \\ 2x-y=1 .....\left(\mathrm{ii}\right) \end{gathered}$$
Adding (i) and (ii), we get
3x = 3
⇒ x = 1
Putting x = 1 in (i), we get
1 + y = 2
⇒ y = 1
Thus, the solution of the given equations is (1, 1). 
We know that, infinitely many straight lines pass through a single point.
So, the equation of one such line can be 3x + 2y = 5 or 2x + 3y = 5.

Answer:

The unique solution is given as x = −1 and y = 3.
The one pair of linear equations having x = −1 and y = 3 as unique solution can be
12x + 5y = 3
2x + y = 1
Similarly, infinitely many pairs of linear equations are possible. 

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

By cross multiplication method we get

Also

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

By cross multiplication method we get

So

We know that

Hence we get the value of and

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

T

Thereforeand

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Now rewriting the given equation as

By cross multiplication method we get

Consider the following for u

Consider the following for v

We know that

Now adding eq. (3) and (4) we get

And after substituting the value of x in eq. (4) we get

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following to calculate x

And 

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Now for y

Hence we get the value of and

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get 

Hence we get the value of

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

For y

Hence we get the value of and

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And 

Hence we get the value of and

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And 

Hence we get the value ofand

Answer:

GIVEN: 

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

Let

Rewriting equations 

Now, by cross multiplication method we get

For u consider the following

For y consider

We know that

Now 

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

After rewriting equations

By cross multiplication method we get

For y consider the following

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

And

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now consider the following for y 

Hence we get the value of and

Answer:

GIVEN:

To find: The solution of the systems of equation by the method of cross-multiplication:

Here we have the pair of simultaneous equation

By cross multiplication method we get

Consider the following for x

Now for y

Hence we get the value of and

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has no solution.

Hence the system of equation has no solution

Answer:

GIVEN: 

To find: To determine whether the system has a unique solution, no solution or infinitely many solutions 

We know that the system of equations

For unique solution

For no solution

For infinitely many solution 

Here,

Since which means hence the system of equation has infinitely many solution.

Hence the system of equation has infinitely many solutions

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution.

Answer:

GIVEN: 

To find: To determine to value of k for which the system has a unique solution.

We know that the system of equations

For unique solution

Here,

Hence for the system of equation has unique solution

Answer:

GIVEN: 

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Consider the following for k

Now consider the following

Hence for the system of equation have infinitely many solutions.

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Hence for the system of equation have infinitely many solutions.

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here,

Consider the following relation to find k

Now again consider the following

Hence for the system of equation have infinitely many solutions

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has infinitely many solutions 

We know that the system of equations

For infinitely many solution 

Here

Now again consider the following to find k

Hence for the system of equation have infinitely many solutions

Answer:

GIVEN:

To find: To determine for what value of k the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Answer:

GIVEN:

To find: To determine for what value of c the system of equation has no solution 

We know that the system of equations

For no solution

Here,

Hence for the system of equation has no solution.

Answer:

​kx + 3y = k − 3
12x + ky = k

We know that the system of equations

Has no solution when,

Here,
$$\begin{gathered} a_1=k, b_1=3, c_1=k-3 \\ \mathrm{And} a_2=12, b_2=k, c_2=k \end{gathered}$$

$\frac{k}{12}=\frac{3}{k}\ne \frac{k-3}{k}$
Consider the following for k

$$\begin{gathered} \frac{k}{12}=\frac{3}{k} \\ \Rightarrow k^2=12\times 3 \\ \Rightarrow k^2=36 \\ \Rightarrow k=\pm 6 \end{gathered}$$

Now consider the following

$$\begin{gathered} \frac{3}{k}\ne \frac{k-3}{k} \\ \Rightarrow 3k\ne k(k-3) \\ \Rightarrow 3k\ne k^2-3k \\ \Rightarrow 6k\ne k^2 \\ \Rightarrow k\ne 6 \end{gathered}$$
Thus, the common value of k is −6.

Hence for k = −6 the system of equation has no solution.

Answer:

If the system of equations $a_{1}x+b_{1}y+c_1=0 \mathrm{and} a_{2}x+b_{2}y+c_2=0$ has infinitely many solutions, then $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.
Since, the system of equations cx + 3y + 3 – c = 0, 12x + cy – c = 0 has infinitely many solutions
Therefore,
$$\begin{gathered} \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{c}{12}=\frac{3}{c} \\ \Rightarrow c\times c=3\times 12 \\ \Rightarrow c^2=36 \\ \Rightarrow c=\pm 6 \\ \mathrm{For} c=6, \\ \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{6}{12}=\frac{3}{6}=\frac{3-6}{-6}=\frac{1}{2} \\ \mathrm{Hence}, \mathrm{it} \mathrm{holds} \mathrm{for} c=6. \\ \mathrm{For} c=-6, \\ \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c} \\ \Rightarrow \frac{-6}{12}=\frac{3}{-6}=\frac{3-\left(-6\right)}{6}\ne \frac{-1}{2} \\ \mathrm{Hence}, \mathrm{it} \mathrm{does} \mathrm{not} \mathrm{holds} \mathrm{for} c=-6. \end{gathered}$$

Hence, the value of c is 6.