Rs Aggarwal 2019 2020 for Class 7 Math Chapter 6 - Algebraic Expressions

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Rs Aggarwal 2019 2020 Solutions for Class 7 Math Chapter 6 Algebraic Expressions are provided here with simple step-by-step explanations. These solutions for Algebraic Expressions are extremely popular among Class 7 students for Math Algebraic Expressions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 7 Math Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 99:

Question 1:

Add the following expressions:

(i) 5x, 7x, −6x
(ii) $\frac{3}{5} x, \frac{2}{3} x, \frac{- 4}{5} x$
(iii) 5a²b, −8a²b, 7a²b
(iv) $\frac{3}{4} x^{2}, 5 x^{2}, - 3 x^{2}, - \frac{1}{4} x^{2}$
(v) x − 3y + 4z, y − 2x − 8z, 5x − 2y − 3z
(vi) 2x² − 3y², 5x² + 6y², − 3x² − 4y²
(vii) 5x − 2x² − 8, 8x²− 7x − 9, 3 + 7x² − 2x
(viii) $\frac{2}{3} a - \frac{4}{5} b + \frac{3}{5} c, - \frac{3}{4} a - \frac{5}{2} b + \frac{2}{3} c, \frac{5}{2} a + \frac{7}{4} b - \frac{5}{6} c$
(ix) $\frac{8}{5} x + \frac{11}{7} y + \frac{9}{4} x y, - \frac{3}{2} x - \frac{5}{3} y - \frac{9}{5} x y$
(x) $\frac{3}{2} x^{3} - \frac{1}{4} x^{2} + \frac{5}{3}, - \frac{5}{4} x^{3} + \frac{3}{5} x^{2} - x + \frac{1}{5}, - x^{2} + \frac{3}{8} x - \frac{8}{15}$

Answer:

(i)
5x + 7x + (-6x)
= 5x + 7x -6x
= 6x

(ii)

$\frac{3}{5} x + \frac{2}{3} x + \frac{- 4}{5} x = \frac{9 x + 10 x - 12 x}{15} = \frac{7 x}{15}$

(iii)
5a²b +( −8a²b) + 7a²b
= 5a²b − 8a²b + 7a²b
4a²b

(iv)
$\frac{3}{4} x^{2} + 5 x^{2} + (- 3 x^{2}) + (- \frac{1}{4} x^{2}) = \frac{3}{4} x^{2} - \frac{1}{4} x^{2} + 5 x^{2} - 3 x^{2} = \frac{1}{2} x^{2} + 2 x^{2} = \frac{5}{2} x^{2}$

(v) Collecting like terms and adding them:

x − 3y + 4z + y − 2x − 8z + 5x − 2y − 3z
= x- 2x + 5x - 3y + y - 2y + 4z - 8z - 3z
= 4x -4y -7z

(vi) Collecting like terms and adding them:
2x² − 3y² + 5x² + 6y² + (− 3x² − 4y²)
$= 2 x^{2} + 5 x^{2} - 3 x^{2} - 3 y^{2} + 6 y^{2} - 4 y^{2} = 4 x^{2} - y^{2}$

(vii) Collecting like terms and adding them:
5x − 2x² − 8 + 8x²− 7x − 9 + 3 + 7x² − 2x
$= - 2 x^{2} + 8 x^{2} + 7 x^{2} + 5 x - 7 x - 2 x - 8 - 9 + 3 = 13 x^{2} - 4 x - 14$

(viii) Collecting like terms and adding them:

$\frac{2}{3} a - \frac{4}{5} b + \frac{3}{5} c + (- \frac{3}{4} a - \frac{5}{2} b + \frac{2}{3} c) + \frac{5}{2} a + \frac{7}{4} b - \frac{5}{6} c b - \frac{5}{2} b + \frac{7}{4} b + \frac{3}{5} c + \frac{2}{3} c - \frac{5}{6} c = \frac{(8 - 9 + 30) a}{12} + \frac{(- 16 - 50 + 35) b}{20} + \frac{(18 + 20 - 25) c}{30} = \frac{29}{12} a - \frac{31}{20} b + \frac{13}{30} c$

(ix) Collecting like terms and adding them:
$\frac{8}{5} x + \frac{11}{7} y + \frac{9}{4} xy + (- \frac{3}{2} x - \frac{5}{3} y - \frac{9}{5} xy) = \frac{8}{5} x - \frac{3}{2} x + \frac{11}{7} y - \frac{5}{3} y + \frac{9}{4} xy - \frac{9}{5} xy = \frac{1}{10} x - \frac{2}{21} y + \frac{9}{20} xy$

(x) Collecting like terms and adding them:
$\frac{3}{2} x^{3} - \frac{1}{4} x^{2} + \frac{5}{3} + (- \frac{5}{4} x^{3} + \frac{3}{5} x^{2} - x + \frac{1}{5}) + (- x^{2} + \frac{3}{8} x - \frac{8}{15}) = \frac{3}{2} x^{3} - \frac{5}{4} x^{3} - \frac{1}{4} x^{2} + \frac{3}{5} x^{2} - x^{2} - x + \frac{3}{8} x + \frac{5}{3} + \frac{1}{5} - \frac{8}{15} = \frac{1}{4} x^{3} - \frac{13}{20} x^{2} - \frac{5}{8} x + \frac{4}{3}$ x+117y+94xy,−32x−53y−9

Page No 99:

Question 2:

Subtract:

(i) −8xy from 7xy
(ii) x² from − 3x²
(iii) (x − y) from (4y − 5x)
(iv) (a²+ b² − 2ab) from (a² + b² + 2ab)
(v) (x² − y²) from (2x² − 3y² + 6xy)
(vi) (x − y + 3z) from (2z − x − 3y)

Answer:

(i) 7xy- (-8xy)
= 7xy+ 8xy
= 15xy

(ii) - 3x² - x²
= -4x²

^{(iii) (4y - 5x) - (x- y)

= 4y - 5x - x + y

= 5y - 6x

(iv)}(a² + b² + 2ab) - (a²+ b² − 2ab)
= $a^{2} - a^{2} + b^{2} - b^{2} + 2 a b + 2 a b$     (Collecting like terms and adding them)
= 4ab

(v) (2x² − 3y² + 6xy) - (x² − y²)
$2 x^{2} - x^{2} - 3 y^{2} + y^{2} + 6 x y = x^{2} - 2 y^{2} + 6 x y$      (Collecting like terms and adding them)

(vi) (2z -x -3y) - (x - y +3z)
= 2z -3z -x -x -3y +y       (Collecting like terms and adding them)
= -z -2x - 2y

Page No 99:

Question 3:

Subtract (2a − 3b + 4c) from the sum of (a + 3b − 4c), (4a − b + 9c) and (−2b + 3c − a).

Answer:

(a + 3b − 4c) + (4a − b + 9c) + (−2b + 3c − a)
= a + 4a - a + 3b -b -2b -4c +9c + 3c
= 4a + 8c

Now, (4a + 8c ) - (2a − 3b + 4c)
= 4a - 2a + 3b + 8c - 4c
= 2a + 3b + 4c

Page No 99:

Question 4:

Subtract the sum of (8m − 7n + 6p²) and (−3m − 4n − p²) from the sum of (2m + 4n − 3p²) and (− m − n − p²).

Answer:

(8m − 7n + 6p²) + (−3m − 4n − p²)
$= 8 m - 3 m - 7 n - 4 n + 6 p^{2} - p^{2} = 5 m - 11 n + 5 p^{2}$

(2m + 4n − 3p²) + (− m − n − p²).
$= 2 m - m + 4 n - n - 3 p^{2} - p^{2} = m + 3 n - 4 p^{2}$

$Now, (m + 3 n - 4 p^{2}) - (5 m - 11 n + 5 p^{2}) = - 4 m + 14 n - 9 p^{2}$

Page No 99:

Question 5:

Subtract the sum of (8a − 6a² + 9) and (−10a − 8 + 8a²) from −3.

Answer:

(8a − 6a² + 9)+ (−10a − 8 + 8a²)

Collecting like terms and adding them:

$8 a - 10 a - 6 a^{2} + 8 a^{2} + 9 - 8 = - 2 a + 2 a^{2} + 1 Now, - 3 - (- 2 a + 2 a^{2} + 1) = 2 a - 2 a^{2} - 4$

Page No 100:

Question 6:

Simplify:

(i) (5x −9y) − (−7x + y)
(ii) $(x^{2} - x) - \frac{1}{2} (x - 3 + 3 x^{2})$
(iii) [7 − 2x + 5y − (x − y)] − (5x + 3y − 7)
(iv) $(\frac{1}{3} y^{2} - \frac{4}{7} y + 5) - (\frac{2}{7} y - \frac{2}{3} y^{2} + 2) - (\frac{1}{7} y - 3 + 2 y^{2})$

Answer:

Collecting like terms and adding them:

(i) 5x + 7x - 9y -y
= 12x -10y

(ii)
$x^{2} - \frac{3}{2} x^{2} - x - \frac{1}{2} x + \frac{3}{2} = - \frac{1}{2} x^{2} - \frac{3}{2} x + \frac{3}{2}$

(iii) 7 + 7 - 2x -x - 5x + 5y + y - 3y
= 14 - 8x -3y

(iv)
$\frac{1}{3} y^{2} + \frac{2}{3} y^{2} - 2 y^{2} - \frac{4}{7} y - \frac{2}{7} y - \frac{1}{7} y + 5 - 2 + 3 = - y^{2} - y + 6$

Page No 102:

Question 1:

Find the products:

3a² × 8a⁴

Answer:

3a² × 8a⁴
^{$= (3 \times 8) \times (a^{2} \times a^{4}) = 24 \times a^{(2 + 4)} = 24 a^{6}$}

Page No 102:

Question 2:

Find the products:

−6x³ × 5x²

Answer:

−6x³ × 5x²
^{$= (- 6 \times 5) \times (x^{3} \times x^{2}) = (- 30) \times (x^{(3 + 2)}) = - 30 x^{5}$}

Page No 102:

Question 3:

Find the products:

(−4ab) × (−3a²bc)

Answer:

(−4ab) × (−3a²bc)
$= (- 4 \times - 3) \times (a \times a^{2} \times b \times b \times c) = 12 \times (a^{3} b^{2} c) = 12 a^{3} b^{2} c$

Page No 102:

Question 4:

Find the products:

(2a²b³) × (−3a³b)

Answer:

(2a²b³) × (−3a³b)
$= (2 \times (- 3)) \times (a^{2} \times a^{3} \times b^{3} \times b) = (- 6) \times (a^{(2 + 3)} \times b^{(3 + 1)}) = - 6 a^{5} b^{4}$

Page No 102:

Question 5:

Find the products:

$\frac{2}{3} x^{2} y \times \frac{3}{5} x y^{2}$

Answer:

$= (\frac{2}{3} \times \frac{3}{5}) \times (x^{2} \times x \times y \times y^{2)}) = \frac{2}{5} \times x^{(2 + 1)} \times y^{(1 + 2)} = \frac{2}{5} x^{3} y^{3}$

Page No 102:

Question 6:

Find the products:

$(\frac{- 3}{4} a b^{3}) \times (\frac{- 2}{3} a^{2} b^{4})$

Answer:

$= (\frac{- 3}{4} \times \frac{- 2}{3}) \times (a \times a^{2} \times b^{3} \times b^{4}) = \frac{1}{2} \times a^{(1 + 2)} \times b^{(3 + 4)} = \frac{1}{2} a^{3} b^{7}$

Page No 102:

Question 7:

Find the products:

$(\frac{- 1}{27} a^{2} b^{2}) \times (\frac{- 9}{2} a^{3} b c^{2})$

Answer:

$= (\frac{- 1}{27} \times \frac{- 9}{2}) \times (a^{2} \times a^{3} \times b^{2} \times b \times c^{2}) = \frac{1}{6} \times a^{(2 + 3)} \times b^{(2 + 1)} \times c^{2} = \frac{1}{6} a^{5} b^{3} c^{2}$

Page No 102:

Question 8:

Find the products:

$(\frac{- 13}{5} a b^{2} c) \times (\frac{7}{3} a^{2} b c^{2})$

Answer:

$= (\frac{- 13}{5} \times \frac{7}{3}) \times (a \times a^{2} \times b^{2} \times b \times c \times c^{2}) = \frac{- 91}{15} a^{(1 + 2)} \times b^{(2 + 1)} \times c^{(1 + 2)} = \frac{- 91}{15} a^{3} b^{3} c^{3}$

Page No 102:

Question 9:

Find the products:

$(\frac{- 18}{5} x^{2} z) \times (\frac{- 25}{6} x z^{2} y)$

Answer:

$= (- \frac{18}{5} \times \frac{- 25}{6}) \times (x^{2} \times x \times z \times z^{2} \times y) = 15 \times x^{(2 + 1)} \times y \times z^{(1 + 2)} = 15 x^{3} {yz}^{3}$

Page No 102:

Question 10:

Find the products:

$(\frac{- 3}{14} x y^{4}) \times (\frac{7}{6} x^{3} y)$

Answer:

$= (\frac{- 3}{14} \times \frac{7}{6}) \times (x \times x^{3} \times y^{4} \times y) = \frac{- 1}{4} x^{(1 + 3)} \times y^{(4 + 1)} = \frac{- 1}{4} x^{4} y^{5}$

Page No 102:

Question 11:

Find the products:

$(\frac{- 7}{5} x^{2} y) \times (\frac{3}{2} x y^{2}) \times (\frac{- 6}{5} x^{3} y^{3})$

Answer:

$= (\frac{- 7}{5} \times \frac{3}{2} \times \frac{- 6}{5}) \times (x^{2} \times x \times x^{3} \times y \times y^{2} \times y^{3}) = \frac{63}{25} \times x^{(2 + 1 + 3)} \times y^{(1 + 2 + 3)} = \frac{63}{25} x^{6} y^{6}$

Page No 102:

Question 12:

Find the products:

(2a²b) × (−5ab²c) × (−6bc²)

Answer:

$= (2 \times (- 5) \times (- 6)) \times (a^{2} \times a \times b \times b^{2} \times b \times c \times c^{2}) = 60 \times a^{(2 + 1)} \times b^{(1 + 2 + 1)} \times c^{(1 + 2)} = 60 a^{3} b^{4} c^{3}$

Page No 102:

Question 13:

Find the products:

(−4x²) × (−6xy²) × (−3y)

Answer:

$= (- 4 \times (- 6) \times (- 3)) \times (x^{2} \times x \times y^{2} \times y) = - 72 \times x^{(2 + 1)} \times y^{(2 + 1)} = - 72 x^{3} y^{3}$

Page No 102:

Question 14:

Find the products:

$(\frac{- 3}{5} s^{2} t) \times (\frac{15}{7} s t^{2} u) \times (\frac{7}{9} s u^{2})$

Answer:

$= (\frac{- 3}{5} \times \frac{15}{7} \times \frac{7}{9}) \times (s^{2} \times s \times s \times t \times t^{2} \times u \times u^{2}) = - 1 \times s^{(2 + 1 + 1)} \times t^{(1 + 2)} \times u^{(1 + 2)} = - s^{4} t^{3} u^{3}$

Page No 102:

Question 15:

Find the products:

$(\frac{- 2}{7} u^{4} v) \times (\frac{- 14}{5} u v^{3}) \times (\frac{- 3}{4} u^{2} v^{3})$

Answer:

$= (\frac{- 2}{7} \times \frac{- 14}{5} \times \frac{- 3}{4}) \times (u^{4} \times u \times u^{2} \times v \times v^{3} \times v^{3}) = \frac{- 3}{5} \times u^{(4 + 1 + 2)} \times v^{(1 + 3 + 3)} = \frac{- 3}{5} u^{7} v^{7}$

Page No 102:

Question 16:

Find the products:

(ab²) × (−b²c) × (−a²c³) × (−3abc)

Answer:

$= (- 3 \times - 1 \times - 1) \times (a \times a^{2} \times a \times b^{2} \times b^{2} \times b \times c \times c^{3} \times c = - 3 \times a^{(1 + 2 + 1)} \times b^{(2 + 2 + 1)} \times c^{(1 + 4 + 1)} = - 3 a^{4} b^{5} c^{5}$

Page No 102:

Question 17:

Find the products:

$(\frac{4}{3} x^{2} y z) \times (\frac{1}{3} y^{2} z x) \times (- 6 x y z^{2})$

Answer:

$= (\frac{4}{3} \times \frac{1}{3} \times (- 6)) \times (x^{2} \times x \times x \times y \times y^{2} \times y \times z \times z \times z^{2}) = \frac{- 8}{3} \times x^{(2 + 1 + 1)} \times y^{(1 + 2 + 1)} \times z^{(1 + 1 + 2)} = \frac{- 8}{3} x^{4} y^{4} z^{4}$

Page No 102:

Question 18:

Multiply $- \frac{2}{3} a^{2} b by \frac{6}{5} a^{3} b^{2}$ and verify your result for a = 2 and b = 3.

Answer:

$\frac{- 2}{3} a^{2} b \times \frac{6}{5} a^{3} b^{2} = (\frac{- 2}{3} \times \frac{6}{5}) \times (a^{2} \times a^{3} \times b \times b^{2}) = \frac{- 4}{5} \times a^{(2 + 3)} \times b^{(1 + 2)} = \frac{- 4}{5} a^{5} b^{3}$

When a =2 and b =3, we get:

$\frac{- 2}{3} a^{2} b = \frac{- 2}{3} \times 2^{2} \times 3 = - 8 \frac{6}{5} a^{3} b^{2} = \frac{6}{5} \times 2^{3} \times 3^{2} = \frac{432}{5} L . H . S . = \frac{- 2}{3} a^{2} b \times \frac{6}{5} a^{3} b^{2} = - 8 \times \frac{432}{5} = \frac{- 3456}{5} R . H . S . = \frac{- 4}{5} a^{5} b^{3} = \frac{- 4}{5} \times 2^{5} \times 3^{3} = \frac{- 3456}{5}$

L.H.S. = R.H.S.

Hence, the result is verified.

Page No 102:

Question 19:

Multiply $- \frac{8}{21} x^{2} y^{3} by - \frac{7}{16} x y^{2}$ and verify your result for x = 3 and y = 2.

Answer:

$\frac{- 8}{21} x^{2} y^{3} \times \frac{- 7}{16} {xy}^{2} = (\frac{- 8}{21} \times \frac{- 7}{16}) (x^{2 + 1}) (y^{3 + 2}) = \frac{1}{6} \times x^{3} \times y^{5} When x = 3 and y = 2, we get : L . H . S . = \frac{- 8}{21} x^{2} y^{3} \times \frac{- 7}{16} {xy}^{2} = \frac{- 192}{7} \times \frac{- 21}{4} = 144 R . H . S . = \frac{1}{6} x^{3} y^{5} = \frac{1}{6} \times 3^{3} \times 2^{5} = 144 L . H . S . = R . H . S . ∴ \frac{- 8}{21} x^{2} y^{3} \times \frac{- 7}{16} {xy}^{2} = \frac{1}{6} x^{3} y^{5}$

Page No 102:

Question 20:

Find the value of (2.3a⁵b²) × (1.2a²b²), when a = 1 and b = 0.5.

Answer:

$= (2.3 \times 1.2) \times (a^{5} \times a^{2} \times b^{2} \times b^{2}) = 2.76 \times a^{(5 + 2)} \times b^{(2 + 2)} = 2.76 a^{7} b^{4} When a = 1 and b = 0.5, we get : 2.76 a^{7} b^{4} = 2.76 \times 1^{7} \times 0 . 5^{4} = 0.1725$

Page No 102:

Question 21:

Find the value of (−8u²v⁶) × (−20uv) for u = 2.5 and v = 1.

Answer:

$= (- 8 \times (- 20)) \times (u^{2} \times u \times v^{6} \times v) = 160 \times u^{(2 + 1)} \times v^{(6 + 1)} = 160 u^{3} v^{7} 160 u^{3} v^{7} = 160 \times 2 . 5^{3} \times 1^{7} = 2500$

Page No 103:

Question 22:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$(\frac{2}{5} a^{2} b) \times (- 15 b^{2} a c) \times (- \frac{1}{2} c^{2})$

Answer:

$= (\frac{2}{5} \times - 15 \times \frac{- 1}{2}) \times (a^{2} \times a \times b \times b^{2} \times c \times c^{2}) = 3 \times a^{(2 + 1)} \times b^{(1 + 2)} \times c^{(1 + 2)} = 3 a^{3} b^{3} c^{3} When a = 1, b = 2 and c = 3, we get : \frac{2}{5} a^{2} b = \frac{2}{5} \times 1^{2} \times 2 = \frac{4}{5} - 15 b^{2} ac = - 15 \times 2^{2} \times 1 \times 3 = - 180 \frac{- 1}{2} c^{2} = \frac{- 1}{2} \times 3^{2} = \frac{- 9}{2} L . H . S . = \frac{2}{5} a^{2} b \times - 15 b^{2} ac \times \frac{- 1}{2} c^{2} = \frac{4}{5} \times - 180 \times \frac{- 9}{2} = 648 R . H . S . = 3 a^{3} b^{3} c^{3} = 3 \times 1^{3} \times 2^{3} \times 3^{3} = 648 L . H . S . = R . H . S . ∴ \frac{2}{5} a^{2} b \times - 15 b^{2} ac \times \frac{- 1}{2} c^{2} = 3 a^{3} b^{3} c^{3}$

Page No 103:

Question 23:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$(\frac{1}{4} a b c) \times (- 6 b^{2} c) \times (- \frac{1}{3} c^{3})$

Answer:

$= (\frac{1}{4} \times - 6 \times - \frac{1}{3}) \times (a \times b \times b^{2} \times c \times c \times c^{3}) = \frac{1}{2} \times a \times b^{(1 + 2)} \times c^{(1 + 1 + 3)} = \frac{1}{2} {ab}^{3} c^{5} When a = 1, b = 2 and c = 3, we get : \frac{1}{4} abc = \frac{1}{4} \times 1 \times 2 \times 3 = \frac{3}{2} - 6 b^{2} c = - 6 \times 2^{2} \times 3 = - 72 - \frac{1}{3} c^{3} = \frac{- 1}{3} \times 3^{3} = - 9 L . H . S . = \frac{1}{4} abc \times - 6 b^{2} c \times - \frac{1}{3} c^{3} = \frac{3}{2} \times - 72 \times - 9 = 972 R . H . S . = \frac{1}{2} {ab}^{3} c^{5} = \frac{1}{2} \times 1 \times 2^{3} \times 3^{5} = 972 L . H . S . = R . H . S . ∴ \frac{1}{4} abc \times - 6 b^{2} c \times - \frac{1}{3} c^{3} = \frac{1}{2} {ab}^{3} c^{5}$

Page No 103:

Question 24:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$(\frac{4}{9} a b c^{3}) \times (\frac{- 27}{5} a^{3} b^{2}) \times (- 8 b^{3} c)$

Answer:

$= (\frac{4}{9} \times \frac{- 27}{5} \times - 8) \times (a \times a^{3} \times b \times b^{2} \times b^{3} \times c^{3} \times c) = \frac{96}{5} \times a^{(1 + 3)} \times b^{(1 + 2 + 3)} \times c^{(3 + 1)} = \frac{96}{5} a^{4} b^{6} c^{4} When a = 1, b = 2 and c = 3 : L . H . S . : (\frac{4}{9} \times \frac{- 27}{5} \times - 8) \times (a \times a^{3} \times b \times b^{2} \times b^{3} \times c^{3} \times c) = (\frac{4}{9} \times \frac{- 27}{5} \times - 8) \times (1 \times 1^{3} \times 2 \times 2^{2} \times 2^{3} \times 3^{3} \times 3) = \frac{497664}{5} R . H . S . : \frac{96}{5} a^{4} b^{6} c^{4} = \frac{96}{5} (1^{4} \times 2^{6} \times 3^{4}) = \frac{497664}{5} L . H . S . = R . H . S . Hence, verified .$

Page No 103:

Question 25:

Find the product and verify the result for a = 1, b = 2 and c = 3.

$(\frac{- 4}{7} a^{2} b) \times (\frac{- 2}{3} b^{2} c) \times (\frac{- 7}{6} c^{2} a)$

Answer:

$= (\frac{- 4}{7} \times \frac{- 2}{3} \times \frac{- 7}{6}) \times (a^{2} \times a \times b \times b^{2} \times c \times c^{2}) = - \frac{4}{9} a^{(2 + 1)} \times b^{(1 + 2)} \times c^{(1 + 2)} = \frac{- 4}{9} a^{3} b^{3} c^{3} L . H . S . : (\frac{- 4}{7} \times \frac{- 2}{3} \times \frac{- 7}{6}) \times (1^{2} \times 1 \times 2 \times 2^{2} \times 3 \times 3^{2}) = - 96 R . H . S . : \frac{- 4}{9} \times 1^{3} \times 2^{3} \times 3^{3} = - 96 L . H . S . = R . H . S . Hence, verified .$

Page No 104:

Question 1:

Find the product:

4a(3a + 7b)

Answer:

$= 4 a \times 3 a + 4 a \times 7 b = 4 \times 3 \times a^{(1 + 1)} + 4 \times 7 \times a \times b = 12 a^{2} + 28 ab$

Page No 104:

Question 2:

Find the product:

5a(6a − 3b)

Answer:

$= 5 a \times 6 a - 5 a \times 3 b = 5 \times 6 \times a \times a - (5 \times 3 \times a \times b) = 30 a^{2} - 15 ab$

Page No 104:

Question 3:

Find the product:

8a²(2a + 5b)

Answer:

$= 8 a^{2} \times 2 a + 8 a^{2} \times 5 b = 8 \times 2 \times a^{2} \times a + 8 \times 5 \times a^{2} \times b = 16 a^{(2 + 1)} + 40 a^{2} b = 16 a^{3} + 40 a^{2} b$

Page No 104:

Question 4:

Find the product:

9x²(5x + 7)

Answer:

$= 9 x^{2} \times 5 x + 9 x^{2} \times 7 = 9 \times 5 \times x^{2} \times x + 9 \times 7 \times x^{2} = 45 x^{(2 + 1)} + 63 x^{2} = 45 x^{3} + 63 x^{2}$

Page No 104:

Question 5:

Find the product:

ab(a²− b²)

Answer:

$= ab \times a^{2} - ab \times b^{2} = a^{(1 + 2)} b - {ab}^{(1 + 2)} = a^{3} b - {ab}^{3}$

Page No 104:

Question 6:

Find the product:

2x²(3x − 4x²)

Answer:

$= 2 x^{2} \times 3 x - 2 x^{2} \times 4 x^{2} = 2 \times 3 \times x^{2} \times x - 2 \times 4 \times x^{2} \times x^{2} = 6 \times x^{(2 + 1)} - 8 \times x^{(2 + 2)} = 6 x^{3} - 8 x^{4}$

Page No 104:

Question 7:

Find the product:

$\frac{3}{5} m^{2} n (m + 5 n)$

Answer:

$= \frac{3}{5} m^{2} n \times m + \frac{3}{5} m^{2} n \times 5 n = \frac{3}{5} \times m^{2} \times m \times n + \frac{3}{5} \times 5 \times m^{2} \times n \times n = \frac{3}{5} m^{(2 + 1)} \times n + 3 \times m^{2} \times n^{(1 + 1)} = \frac{3}{5} m^{3} n + 3 m^{2} n^{2}$

Page No 104:

Question 8:

Find the product:

−17x²(3x − 4)

Answer:

$= - 17 x^{2} \times 3 x - (- 17 x^{2} \times 4) = - 17 \times 3 \times x^{2} \times x + 17 \times 4 \times x^{2} = - 51 \times x^{(2 + 1)} + 68 x^{2} = - 51 x^{3} + 68 x^{2}$

Page No 104:

Question 9:

Find the product:

$\frac{7}{2} x^{2} (\frac{4}{7} x + 2)$

Answer:

$= \frac{7}{2} x^{2} \times \frac{4}{7} \times x + \frac{7}{2} x^{2} \times 2 = \frac{7}{2} \times \frac{4}{7} \times x^{2} \times x + \frac{7}{2} \times 2 \times x^{2} = 2 \times x^{(2 + 1)} + 7 x^{2} = 2 x^{3} + 7 x^{2}$

Page No 104:

Question 10:

Find the product:

−4x²y(3x²− 5y)

Answer:

$= - 4 x^{2} y \times 3 x^{2} - (- 4 x^{2} y \times 5 y) = - 4 \times 3 \times x^{2} \times x^{2} \times y + 4 \times 5 \times x^{2} \times y \times y = - 12 \times x^{(2 + 2)} \times y + 20 \times x^{2} \times y^{(1 + 1)} = - 12 x^{4} y + 20 x^{2} y^{2}$

Page No 104:

Question 11:

Find the product:

$\frac{- 4}{27} x y z (\frac{9}{2} x^{2} y z - \frac{3}{4} x y z^{2})$

Answer:

$= \frac{- 4}{27} xyz \times \frac{9}{2} x^{2} yz - (\frac{- 4}{27} xyz \times \frac{3}{4} {xyz}^{2}) = \frac{- 4}{27} \times \frac{9}{2} \times x \times x^{2} \times y \times y \times z \times z + \frac{4}{27} \times \frac{3}{4} \times x \times x \times y \times y \times z \times z^{2} = \frac{- 2}{3} \times x^{(1 + 2)} \times y^{(1 + 1)} \times z^{(1 + 1)} + \frac{1}{9} \times x^{(1 + 1)} \times y^{(1 + 1)} \times z^{(1 + 2)} = \frac{- 2}{3} x^{3} y^{2} z^{2} + \frac{1}{9} x^{2} y^{2} z^{3}$

Page No 104:

Question 12:

Find the product:

9t²(t + 7t³)

Answer:

$= 9 t^{2} \times t + 9 t^{2} \times 7 t^{3} = 9 \times t^{2} \times t + 9 \times 7 \times t^{2} \times t^{3} = 9 \times t^{(2 + 1)} + 63 \times t^{(2 + 3)} = 9 t^{3} + 63 t^{5}$

Page No 104:

Question 13:

Find the product:

10a²(0.1a − 0.5b)

Answer:

$= 10 a^{2} \times 0.1 a - 10 a^{2} \times 0.5 b = 10 \times 0.1 \times a^{2} \times a - 10 \times 0.5 \times a^{2} \times b = 1 \times a^{(2 + 1)} - 5 a^{2} b = a^{3} - 5 a^{2} b$

Page No 104:

Question 14:

Find the product:

1.5a(10a²− 100ab²)

Answer:

$= 1.5 a \times 10 a^{2} b - 1.5 a \times 100 {ab}^{2} = 1.5 \times 10 \times a \times a^{2} b - 1.5 \times 100 \times a \times a \times b^{2} = 15 \times a^{(1 + 2)} b - 150 \times a^{(1 + 1)} \times b^{2} = 15 a^{3} b - 150 a^{2} b^{2}$

Page No 104:

Question 15:

Find the product:

$\frac{2}{3} a b c (a^{2} + b^{2} - 3 c^{2})$

Answer:

$= \frac{2}{3} abc \times a^{2} + \frac{2}{3} abc \times b^{2} - \frac{2}{3} abc \times 3 c^{2} = \frac{2}{3} a \times a^{2} \times b \times c + \frac{2}{3} a \times b \times b^{2} \times c - \frac{2}{3} \times 3 \times a \times b \times c \times c^{2} = \frac{2}{3} \times a^{(1 + 2)} \times b \times c + \frac{2}{3} \times a \times b^{(1 + 2)} \times c - 2 \times a \times b \times c^{(1 + 2)} = \frac{2}{3} a^{3} bc + \frac{2}{3} {ab}^{3} c - 2 {abc}^{3}$

Page No 104:

Question 16:

Find the product 24x²(1−2x) and evaluate it for x = 2.

Answer:

${24x}^{2} (1-2x) = 24 x^{2} \times 1 - 24 x^{2} \times 2 x = 24 x^{2} - 24 \times 2 \times x^{2} \times x = 24 x^{2} - 48 x^{3} W h e n x = 2 : L . H . S . = 24 x^{2} (1 - 2 x) = 24 \times 2^{2} (1 - 2 \times 2) = 96 (1 - 4) = 96 \times (- 3) = - 288 R . H . S . = 24 x^{2} - 48 x^{2} = 24 \times 2^{2} - 48 \times 2^{3} = 96 - 384 = - 288 L . H . S . = R . H . S . ∴ 24 x^{2} (1 - 2 x) = 24 x^{2} - 48 x^{3}$

Page No 104:

Question 17:

Find the product ab(a²+b²) and evaluate it for a = 2 and b = $\frac{1}{2}$ .

Answer:

$ab (^{2} {+b}^{2}) = a b \times a^{2} + a b \times b^{2} = a \times a^{2} \times b + a \times b \times b^{2} = a^{(1 + 2)} \times b + a \times b^{(1 + 2)} = a^{3} b + a b^{3} W h e n a = 2 a n d b = \frac{1}{2}, w e g e t : L . H . S . = a b (a^{2} + b^{2}) = 2 \times \frac{1}{2} (2^{2} + \frac{1}{2^{2}}) = 4 + \frac{1}{4} = \frac{17}{4} R . H . S . = a^{3} b + a b^{3} = 2^{3} \times \frac{1}{2} + 2 \times {(\frac{1}{2})}^{3} = 4 + \frac{1}{4} = \frac{17}{4} ∴ L . H . S . = R . H . S .$

Page No 104:

Question 18:

Find the product s (s²− st) and find its value for s = 2 and t = 3.

Answer:

${s (s}^{2} - st) = s \times s^{2} - s \times st = s^{(1 + 2)} - s^{(1 + 1)} \times t = s^{3} - s^{2} t When s = 2 and t = 3, we get : L . H . S . = s (s^{2} - st) = 2 (2^{2} - 2 \times 3) = 2 \times (4 - 6) = - 4 R . H . S . = s^{3} - s^{2} t = 2^{3} - 2^{2} \times 3 = 8 - 12 = - 4 L . H . S . = R . H . S . ∴ s (s^{2} - st) = s^{3} - s^{2} t$

Page No 104:

Question 19:

Find the product −3y(xy + y²) and find its value for x = 4 and y = 5.

Answer:

$- 3 y (xy + y^{2}) = - 3 y \times xy - 3 y \times y^{2} = - 3 \times x \times y \times y - 3 \times y \times y^{2} = - 3 \times x \times y^{(1 + 1)} - 3 \times y^{(1 + 2)} = - 3 {xy}^{2} - 3 y^{3} When x = 4 and y = 5, we get : L . H . S . = - 3 y (xy + y^{2}) = - 3 \times 5 (4 \times 5 + 5^{2}) = - 15 \times (20 + 25) = - 675 R . H . S . = - 3 {xy}^{2} - 3 y^{3} = - 3 \times 4 \times 5^{2} - 3 \times 5^{3} = - 300 - 375 = - 675 L . H . S . = R . H . S . ∴ - 3 y (xy + y^{2}) = - 3 {xy}^{2} - 3 y^{3}$

Page No 104:

Question 20:

Simplify

a(b − c) + b(c − a) + c(a − b)

Answer:

$a(b − c) + b(c − a) + c(a − b) = a \times b - a \times c + b \times c - b \times a + c \times a - c \times b = a b - a c + b c - a b + a c - b c = 0$

Page No 104:

Question 21:

Simplify

a(b − c) − b(c − a) − c(a − b)

Answer:

$a (b - c) - b (c - a) - c (a - b) = a \times b - a \times c - b \times c + b \times a - c \times a + c \times b = ab + ab - ac - ac - bc + bc = 2 ab - 2 ac = 2 a (b - c)$

Page No 104:

Question 22:

Simplify

3x² + 2(x + 2) −3x(2x + 1)

Answer:

$3 x^{2} + 2 (x + 2) - 3 x (2 x + 1) = 3 x^{2} + 2 \times x + 2 \times 2 - 3 x \times 2 x - 3 x = 3 x^{2} + 2 x + 4 - 6 x^{2} - 3 x = - 3 x^{2} - x + 4$

Page No 104:

Question 23:

Simplify

x(x + 4) + 3x(2x² − 1) + 4x² + 4

Answer:

$x (x + 4) + 3 x (2 x^{2} - 1) + 4 x^{2} + 4 = x \times x + x \times 4 + 3 x \times 2 x^{2} - 3 x + 4 x^{2} + 4 = x^{(1 + 1)} + 4 x + 6 \times x^{(1 + 2)} - 3 x + 4 x^{2} + 4 = x^{2} + 4 x + 6 x^{3} - 3 x + 4 x^{2} + 4 = 6 x^{3} + 5 x^{2} + x + 4$

Page No 104:

Question 24:

Simplify

2x² + 3x(1 − 2x³) + x(x + 1)

Answer:

$2 x^{2} + 3 x (1 - 2 x^{3}) + x (x + 1) = 2 x^{2} + 3 x - 3 x \times 2 x^{3} + x^{2} + x = 2 x^{2} + 3 x - 6 \times x^{(1 + 3)} + x^{2} + x = 2 x^{2} + 3 x - 6 x^{4} + x^{2} + x = - 6 x^{4} + 3 x^{2} + 4 x$

Page No 104:

Question 25:

Simplify

a²b(a − b²) + ab²(4ab − 2a²) −a³b(1 − 2b)

Answer:

$a^{2} b (a - b^{2}) + {ab}^{2} (4 ab - 2 a^{2}) - a^{3} b (1 - 2 b) = a^{2} b \times a - a^{2} b \times b^{2} + {ab}^{2} \times 4 ab - {ab}^{2} \times 2 a^{2} - a^{3} b + a^{3} b \times 2 b = a^{(2 + 1)} \times b - a^{2} \times b^{(1 + 2)} + 4 \times a^{(1 + 1)} \times b^{(2 + 1)} - 2 \times a^{(1 + 2)} \times b^{2} - a^{3} b + 2 \times a^{3} \times b^{(1 + 1)} = a^{3} b - a^{2} b^{3} + 4 a^{2} b^{3} - 2 a^{3} b^{2} - a^{3} b + 2 a^{3} b^{2} = 3 a^{2} b^{3}$

Page No 104:

Question 26:

Simplify

4st(s − t) −6s²(t − t²) −3t² (2s²− s) +2st (s − t)

Answer:

$4 st (s - t) - 6 s^{2} (t - t^{2}) - 3 t^{2} (2 s^{2} - s) + 2 st (s - t) = 4 st \times s - 4 st \times t - 6 s^{2} \times t - 6 s^{2} \times (- t^{2}) - 3 t^{2} \times 2 s^{2} - 3 t^{2} \times (- s) + 2 st \times s - 2 st \times t = 4 \times s^{(1 + 1)} \times t - 4 \times s \times t^{(1 + 1)} - 6 s^{2} t + 6 s^{2} t^{2} - 6 t^{2} s^{2} + 3 t^{2} s + 2 \times s^{(1 + 1)} \times t - 2 \times s \times t^{(1 + 1)} = 4 s^{2} t - 4 {st}^{2} - 6 s^{2} t + 6 s^{2} t^{2} - 6 t^{2} s^{2} + 3 t^{2} s + 2 s^{2} t - 2 {st}^{2} = 4 s^{2} t - 6 s^{2} t + 2 s^{2} t - 4 {st}^{2} + 3 {st}^{2} - 2 {st}^{2} = - 3 {st}^{2}$

Page No 106:

Question 1:

Find the product:

(5x + 7)(3x + 4)

Answer:

$(5 x + 7) (3 x + 4) = 5 x \times (3 x + 4) + 7 \times (3 x + 4) = (5 x \times 3 x + 5 x \times 4) + (7 \times 3 x + 7 \times 4) = 15 x^{2} + 20 x + 21 x + 28 = 15 x^{2} + 41 x + 28$

Page No 106:

Question 2:

Find the product:

(4x − 3)(2x + 5)

Answer:

$(4 x - 3) (2 x + 5) = 4 x \times (2 x + 5) - 3 \times (2 x + 5) = (4 x \times 2 x + 4 x \times 5) - (3 \times 2 x + 3 \times 5) = 8 x^{2} + 20 x - 6 x - 15 = 8 x^{2} + 14 x - 15$

Page No 106:

Question 3:

Find the product:

(x − 6)(4x + 9)

Answer:

$(x - 6) (4 x + 9) = x \times (4 x + 9) - 6 \times (4 x + 9) = (x \times 4 x + x \times 9) - (6 \times 4 x + 6 \times 9) = 4 x^{2} + 9 x - 24 x - 54 = 4 x^{2} - 15 x - 54$

Page No 106:

Question 4:

Find the product:

(5y − 1)(3y − 8)

Answer:

$(5 y - 1) (3 y - 8) = 5 y \times (3 y - 8) - 1 \times (3 y - 8) = (5 y \times 3 y - 5 y \times 8) - (3 y - 8) = 15 y^{2} - 40 y - 3 y + 8 = 15 y^{2} - 43 y + 8$

Page No 106:

Question 5:

Find the product:

(7x + 2y)(x + 4y)

Answer:

$(7 x + 2 y) (x + 4 y) = 7 x \times (x + 4 y) + 2 y \times (x + 4 y) = (7 x \times x + 7 x \times 4 y) + (2 y \times x + 2 y \times 4 y) = 7 x^{2} + 28 xy + 2 xy + 8 y^{2} = 7 x^{2} + 30 xy + 8 y^{2}$

Page No 106:

Question 6:

Find the product:

(9x + 5y)(4x + 3y)

Answer:

$(9 x + 5 y) (4 x + 3 y) = 9 x \times (4 x + 3 y) + 5 y \times (4 x + 3 y) = (9 x \times 4 x + 9 x \times 3 y) + (5 y \times 4 x + 5 y \times 3 y) = 36 x^{2} + 27 xy + 20 xy + 15 y^{2} = 36 x^{2} + 47 xy + 15 y^{2}$

Page No 106:

Question 7:

Find the product:

(3m − 4n)(2m − 3n)

Answer:

$(3 m - 4 n) (2 m - 3 n) = 3 m \times (2 m - 3 n) - 4 n \times (2 m - 3 n) = (3 m \times 2 m - 3 m \times 3 n) - (4 n \times 2 m - 4 n \times 3 n) = 6 m^{2} - 9 mn - 8 nm + 12 n^{2} = 6 m^{2} - 17 mn + 12 n^{2}$

Page No 106:

Question 8:

Find the product:

(0.8x − 0.5y)(1.5x − 3y)

Answer:

$(0.8 x - 0.5 y) (1.5 x - 3 y) = 0.8 x \times (1.5 x - 3 y) - 0.5 y \times (1.5 x - 3 y) = (0.8 x \times 1.5 x - 0.8 x \times 3 y) - (0.5 y \times 1.5 x - 0.5 y \times 3 y) = 1.2 x^{2} - 2.4 xy - 0.75 yx + 1.5 y^{2} = 1.2 x^{2} - 3.15 xy + 1.5 y^{2}$

Page No 106:

Question 9:

Find the product:

$(\frac{1}{5} x + 2 y) (\frac{2}{3} x - y)$

Answer:

$(\frac{1}{5} x + 2 y) (\frac{2}{3} x - y) = \frac{1}{5} x \times (\frac{2}{3} x - y) + 2 y \times (\frac{2}{3} x - y) = (\frac{1}{5} x \times \frac{2}{3} x - \frac{1}{5} x \times y) + (2 y \times \frac{2}{3} x - 2 y \times y) = \frac{2}{15} x^{2} - \frac{1}{5} xy + \frac{4}{3} yx - 2 y^{2} = \frac{2}{15} x^{2} + \frac{17}{15} xy - 2 y^{2}$

Page No 106:

Question 10:

Find the product:

$(\frac{2}{5} x - \frac{1}{2} y) (10 x - 8 y)$

Answer:

$(\frac{2}{5} x - \frac{1}{2} y) (10 x - 8 y) = \frac{2}{5} x \times (10 x - 8 y) - \frac{1}{2} y \times (10 x - 8 y) = (\frac{2}{5} x \times 10 x - \frac{2}{5} x \times 8 y) - (\frac{1}{2} y \times 10 x - \frac{1}{2} y \times 8 y) = 4 x^{2} - \frac{16}{5} xy - 5 yx + 4 y^{2} = 4 x^{2} - \frac{41}{5} xy + 4 y^{2}$

Page No 106:

Question 11:

Find the product:

$(\frac{3}{4} a + \frac{2}{3} b) (4 a + 3 b)$

Answer:

$(\frac{3}{4} a + \frac{2}{3} b) (4 a + 3 b) = \frac{3}{4} a \times (4 a + 3 b) + \frac{2}{3} b \times (4 a + 3 b) = (\frac{3}{4} a \times 4 a + \frac{3}{4} a \times 3 b) + (\frac{2}{3} b \times 4 a + \frac{2}{3} b \times 3 b) = 3 a^{2} + \frac{9}{4} ab + \frac{8}{3} ba + 2 b^{2} = 3 a^{2} + \frac{59}{12} ab + 2 b^{2}$

Page No 106:

Question 12:

Find the product:

(x² − a²)(x −a)

Answer:

$(x^{2} - a^{2}) (x - a) = x^{2} \times (x - a) - a^{2} \times (x - a) = (x^{2} \times x - x^{2} \times a) - (a^{2} \times x - a^{2} \times a) = x^{3} - {ax}^{2} - a^{2} x + a^{3}$

Page No 106:

Question 13:

Find the product:

(3p² + q²)(2p² − 3q²)

Answer:

$(3 p^{2} + q^{2} (2 p^{2} - 3 q^{2}) = 3 p^{2} \times (2 p^{2} - 3 q^{2}) + q^{2} \times (2 p^{2} - 3 q^{2}) = (3 p^{2} \times 2 p^{2} - 3 p^{2} \times 3 q^{2}) + (q^{2} \times 2 p^{2} - q^{2} \times 3 q^{2}) = 6 p^{4} - 9 p^{2} q^{2} + 2 q^{2} p^{2} - 3 q^{4} = 6 p^{4} - 7 p^{2} q^{2} - 3 q^{4}$

Page No 106:

Question 14:

Find the product:

(2x² − 5y²)(x² + 3y²)

Answer:

$(2 x^{2} - 5 y^{2}) (x^{2} + 3 y^{2}) = 2 x^{2} \times (x^{2} + 3 y^{2}) - 5 y^{2} \times (x^{2} + 3 y^{2}) = (2 x^{2} \times x^{2} + 2 x^{2} \times 3 y^{2}) - (5 y^{2} \times x^{2} + 5 y^{2} \times 3 y^{2}) = 2 x^{4} + 6 x^{2} y^{2} - 5 y^{2} x^{2} - 15 y^{4} = 2 x^{4} + x^{2} y^{2} - 15 y^{4}$

Page No 106:

Question 15:

Find the product:

(x³ − y³)(x² + y²)

Answer:

$(x^{3} - y^{3}) (x^{2} + y^{2}) = x^{3} \times (x^{2} + y^{2}) - y^{3} \times (x^{2} + y^{2}) = (x^{3} \times x^{2} + x^{3} \times y^{2}) - (y^{3} \times x^{2} + y^{3} \times y^{2}) = x^{5} + x^{3} y^{2} - x^{2} y^{3} - y^{5}$

Page No 106:

Question 16:

Find the product:

(x⁴+ y⁴)(x² − y²)

Answer:

$(x^{4} + y^{4}) (x^{2} - y^{2}) = x^{4} \times (x^{2} - y^{2}) + y^{4} \times (x^{2} - y^{2}) = (x^{4} \times x^{2} - x^{4} \times y^{2}) + (y^{4} \times x^{2} - y^{4} \times y^{2}) = x^{6} - x^{4} y^{2} + x^{2} y^{4} - y^{6}$

Page No 106:

Question 17:

Find the product:

$(x^{4} + \frac{1}{x^{4}}) (x + \frac{1}{x})$

Answer:

$(x^{4} + \frac{1}{x^{4}}) (x + \frac{1}{x}) = x^{4} \times (x + \frac{1}{x}) + \frac{1}{x^{4}} \times (x + \frac{1}{x}) = (x^{4} \times x + x^{4} \times \frac{1}{x}) + (\frac{1}{x^{4}} \times x + \frac{1}{x^{4}} \times \frac{1}{x}) = x^{5} + x^{3} + \frac{1}{x^{3}} + \frac{1}{x^{5}}$

Page No 106:

Question 18:

Find the product:

(x² − y²)(x + 2y)

Answer:

$(x^{2} - y^{2}) (x + 2 y) = x^{2} \times (x + 2 y) - y^{2} \times (x + 2 y) = (x^{2} \times x + x^{2} \times 2 y) - (y^{2} \times x + y^{2} \times 2 y) = x^{3} + 2 x^{2} y - {xy}^{2} - 2 y^{3}$

Page No 106:

Question 19:

Find the product:

(2x + 3y − 5)(x + y)

Answer:

$(2 x + 3 y - 5) (x + y) = 2 x \times (x + y) + 3 y \times (x + y) - 5 \times (x + y) = (2 x \times x + 2 x \times y) + (3 y \times x + 3 y \times y) - (5 \times x + 5 \times y) = 2 x^{2} + 2 x y + 3 y x + 3 y^{2} - 5 x - 5 y = 2 x^{2} + 5 x y - 5 x - 5 y + 3 y^{2}$

Page No 106:

Question 20:

Find the product:

(3x + 2y − 4)(x − y)

Answer:

By column method:

$3 x + 2 y - 4 \times (x - y) 3 x^{2} + 2 yx - 4 x - 3 xy - 2 y^{2} + 4 y Add : 3 x^{2} - xy - 4 x + 4 y - 2 y^{2} ∴ (3 x + 2 y - 4) (x - y) = 3 x^{2} - xy - 4 x + 4 y - 2 y^{2}$

Page No 106:

Question 21:

Find the product:

(x² − 3x + 7)(2x + 3)

Answer:

By column method:

$x^{2} - 3 x + 7 \times (2 x + 3) 2 x^{3} - 6 x^{2} + 14 x 3 x^{2} - 9 x + 21 Add : 2 x^{3} - 3 x^{2} + 5 x + 21 ∴ (x^{2} - 3 x + 7) (2 x + 3) = 2 x^{3} - 3 x^{2} + 5 x + 21$

Page No 106:

Question 22:

Find the product:

(3x² + 5x − 9)(3x −9)

Answer:

By column method:

$3 x^{2} + 5 x - 9 \times (3 x - 9) 9 x^{3} + 15 x^{2} - 27 x - 27 x^{2} - 45 x + 81 Add : 9 x^{3} - 12 x^{2} - 72 x + 81 ∴ (3 x^{2} + 5 x - 9) (3 x - 9) = 9 x^{3} - 12 x^{2} - 72 x + 81$

Page No 106:

Question 23:

Find the product:

(9x² − x + 15)(x² − 3)

Answer:

By column method:

$9 x^{2} - x + 15 \times (x^{2} - 3) 9 x^{4} - x^{3} + 15 x^{2} - 27 x^{2} + 3 x - 45 Add : 9 x^{4} - x^{3} - 12 x^{2} + 3 x - 45 ∴ (9 x^{2} - x + 15) (x^{2} - 3) = 9 x^{4} - x^{3} - 12 x^{2} + 3 x - 45$

Page No 106:

Question 24:

Find the product:

(x²+ xy + y²)(x − y)

Answer:

By column method:

$x^{2} + xy + y^{2} \times (x - y) x^{3} + x^{2} y + {xy}^{2} - x^{2} y - {xy}^{2} - y^{3} Add : x^{3} - y^{3} ∴ (x^{2} + xy + y^{2}) (x - y) = x^{3} - y^{3}$

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Question 25:

Find the product:

(x²− xy + y²)(x + y)

Answer:

By column method:

$x^{2} - xy + y^{2} \times (x + y) x^{3} - x^{2} y + {xy}^{2} x^{2} y - {xy}^{2} + y^{3} Add : x^{3} + y^{3} ∴ (x^{2} - xy + y^{2}) (x + y) = x^{3} + y^{3}$

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Question 26:

Find the product:

(x²− 5x + 8)(x² + 2)

Answer:

By column method:
$x^{2} - 5 x + 8 \times (x^{2} + 2) x^{4} - 5 x^{3} + 8 x^{2} 2 x^{2} - 10 x + 16 Add : x^{4} - 5 x^{3} + 10 x^{2} - 10 x + 16 ∴ (x^{2} - 5 x + 8) (x^{2} + 2) = x^{4} - 5 x^{3} + 10 x^{2} - 10 x + 16$

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Question 27:

Simplify

(3x + 4)(2x − 3) + (5x − 4)(x + 2)

Answer:

(3x +4)(2x -3)
$= 3 x \times (2 x - 3) + 4 \times (2 x - 3) = 6 x^{(1 + 1)} - 9 x + 8 x - 12 = 6 x^{2} - x - 12 (5 x - 4) (x + 2) = 5 x (x + 2) - 4 (x + 2) = 5 x^{(1 + 1)} + 10 x - 4 x - 8 = 5 x^{2} + 6 x - 8$

∴ (3x + 4)(2x − 3) + (5x − 4)(x + 2)

$= 6 x^{2} - x - 12 + 5 x^{2} + 6 x - 8 = 11 x^{2} + 5 x - 20$

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Question 28:

Simplify

(5x − 3)(x + 4) − (2x + 5)(3x − 4)

Answer:

(5x-3)(x+4)

$= 5 x \times (x + 4) - 3 \times (x + 4) = 5 x^{(1 + 1)} + 20 x - 3 x - 12 = 5 x^{2} + 17 x - 12$

(2x +5)(3x-4)

$= 2 x \times (3 x - 4) + 5 \times (3 x - 4) = 6 x^{(1 + 1)} - 8 x + 15 x - 20 = 6 x^{2} + 7 x - 20$

∴ (5x − 3)(x + 4) − (2x + 5)(3x − 4)

$= 5 x^{2} + 17 x - 12 - (6 x^{2} + 7 x - 20) = 5 x^{2} - 6 x^{2} + 17 x - 7 x - 12 + 20 = - x^{2} + 10 x + 8$

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Question 29:

Simplify

(9x − 7)(2x − 5) − (3x − 8)(5x − 3)

Answer:

$(9 x - 7) (2 x - 5) = 9 x \times (2 x - 5) - 7 \times (2 x - 5) = 18 x^{(1 + 1)} - 45 x - 14 x + 35 = 18 x^{2} - 59 x + 35 (3 x - 8) (5 x - 3) = 3 x \times (5 x - 3) - 8 \times (5 x - 3) = 15 x^{2} - 9 x - 40 x + 24 = 15 x^{2} - 49 x + 24 ∴ (2 x − 5) − (3x − 8)(5x − 3) = 18 x^{2} - 59 x + 35 - (15 x^{2} - 49 x + 24) = 18 x^{2} - 15 x^{2} - 59 x + 49 x + 35 - 24 = 3 x^{2} - 10 x + 11$

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Question 30:

Simplify

(2x + 5y)(3x + 4y) − (7x + 3y)(2x − y)

Answer:

(2x +5y)(3x+4y)

$= 2 x \times (3 x + 4 y) + 5 y \times (3 x + 4 y) = 6 x^{(1 + 1)} + 8 xy + 15 yx + 20 y^{(1 + 1)} = 6 x^{2} + 23 xy + 20 y^{2} (7 x + 3 y) (2 x + y) = 7 x (2 x + y) + 3 y (2 x + y) = 14 x^{(1 + 1)} + 7 xy + 6 yx + 3 y^{(1 + 1)} = 14 x^{2} + 13 xy + 3 y^{2}$

∴ (2x + 5y)(3x + 4y) − (7x + 3y)(2x − y)

$= 6 x^{2} + 23 xy + 20 y^{2} - (14 x^{2} + 13 xy + 3 y^{2}) = 6 x^{2} - 14 x^{2} + 23 xy - 13 xy + 20 y^{2} - 3 y^{2} = - 8 x^{2} + 10 xy + 17 y^{2}$

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Question 31:

Simplify

(3x² + 5x − 7)(x − 1) − (x² − 2x + 3)(x + 4)

Answer:

(3x² + 5x − 7)(x − 1)

By column method:

$3 x^{2} + 5 x - 7 \times (x - 1) 3 x^{3} + 5 x^{2} - 7 x - 3 x^{2} - 5 x + 7 Add : 3 x^{3} + 2 x^{2} - 12 x + 7$

(x² − 2x + 3)(x + 4)

By column method:

$x^{2} - 2 x + 3 \times (x + 4) x^{3} - 2 x^{2} + 3 x 4 x^{2} - 8 x + 12 Add : x^{3} + 2 x^{2} - 5 x + 12$

(3x² + 5x − 7)(x − 1) − (x² − 2x + 3)(x + 4)

$= 3 x^{3} + 2 x^{2} - 12 x + 7 - (x^{3} + 2 x^{2} - 5 x + 12) = 3 x^{3} - x^{3} + 2 x^{2} - 2 x^{2} - 12 x + 5 x + 7 - 12 = 2 x^{3} - 7 x - 5$

View NCERT Solutions for all chapters of Class 7