# GSEB Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

Gujarat BoardĀ GSEB Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

Question 1.
Complete the table.

Solution:
We have:
1. a Ć (b + c + d) = a Ć b + a Ć c + a Ć d
= ab + ac + ad

2. (x + y – 5) Ć 5xy
= x Ć 5xy + y Ć 5xy + (-5) Ć 5xy
= 5x2y + 5x2y – 25xy

3. p Ć (6p2 – 7p + 5)
= p Ć 6p2 + p Ć (-7p) + p Ć 5
= 6p3 – 7p2 + 5p

4. 4p2q2 Ć (p2 – q2)
= 4p2q2 Ć p2 + 4p2q2 Ć (-q2)
= (4 Ć 1) Ć p2q2 Ć p2 + [4 x (-1)] Ć p2q2 Ć (-q2)
= 4 Ć p4q2 + (-4) Ć p2q4 = 4p4q2 – 4p2q4

5. (a + b + c) Ć abc
= a Ć abc + b Ć abc + c Ć abc
= (1 Ć 1) Ć a Ć abc + (1 Ć 1) Ć b Ć abc + (1 Ć 1) Ć c Ć abc
= (1) Ć a2bc + (1) Ć ab2c + abc2

Question 2.
Find the product?

1. (a2) Ć (2a22) Ć (4a26)
2. ($$\frac{2}{3}$$xy) Ć ($$\frac{-9}{10}$$x2y2)
3. (-$$\frac{10}{3}$$pq3) Ć ($$\frac{6}{5}$$p3q)
4. x Ć x2 Ć x4 Ć x4

Solution:
1. (a2) Ć (2a22) Ć (4a26)
= (1 Ć 2 Ć 4) Ć a2 Ć a22 Ć a26
= 8 Ć a50 [āµ2 + 22 + 26 = 50]
= 8a50

2. ($$\frac{2}{3}$$xy) Ć (-$$\frac{9}{10}$$x2y2) = $$\frac{2}{3}$$ Ć (-$$\frac{9}{10}$$) Ć xy Ć x2y2
= – $$\frac{2}{3}$$ Ć $$\frac{9}{10}$$ Ć x3y3 = $$\frac{-3}{5}$$x3y3

3. (-$$\frac{10}{3}$$pq3) Ć p3q
= [(-$$\frac{10}{3}$$) Ć $$\frac{6}{5}$$] Ć pq3 Ć p3q
= –$$\frac{10}{3}$$ Ć $$\frac{6}{5}$$ Ć p4q4 = -4p4q4

4. x Ć x2 Ć x3 Ć x4
= (1 Ć 1 Ć 1 Ć 1) Ć x Ć x2 Ć x3 Ć x4
= (1) Ć x10 = x10

Question 3.
(a) Simplify 3x(4x – 5) + 3 and find its values for
(i) x = 3 and
(ii) x = $$\frac{1}{2}$$.

(b) Simplify a(a2 + a + 1) + 5 and find its value for
(i) a = 0
(ii) a = 1 and
(iii) a = -1.
Solution:
(a) 3x(4x – 5) + 3 = 3x Ć 4x + 3x Ć (-5) + 3
= (3 Ć 4) Ć x Ć x + 3 Ć (-5) Ć x + 3
= 12 Ć x2 + (-15) Ć x + 3
= 12x2 – 15x + 3

(i) For x = 3, 12x2 – 15x + 3
= 12(3)2 – 15(3) + 3
= 12 Ć 9 – 45 + 3
= 108 + 3 – 45
= 111 – 45 = 66

(ii) For x = $$\frac{1}{2}$$, 12x2 – 15x + 3
= 12($$\frac{1}{2}$$)2 – 15($$\frac{1}{2}$$) + 3
= 12($$\frac{1}{4}$$) – $$\frac{15}{2}$$ + 3 = 3 – $$\frac{15}{2}$$ + 3 = 6 – $$\frac{15}{2}$$
= $$\frac{12-15}{2}$$ = $$\frac{-3}{2}$$

(b) a(a2 + a + 1) + 5
= a Ć a2 + a Ć a + a Ć 1 + 5
= a3 + a2 + a + 5

(i) For a = 0, a3 + a2 + a + 5
= (1)3 + (1)2 + (1) + 5
= 1 + 1 + 1 + 5 = 8

(ii) For a = 1, a3 + a2 + a + 5
= (1)3 + (1)2 + (1) + 5
= 1 + 1 + 1 + 5 = 8

(iii) For a = -1, a3 + a2 + a + 5
= (-1)3 + (-1)2 + (-1) + 5
= -1 + 1 – 1 + 5 = 4

Question 4.
(a) Add: p(p-q), q(q-r) and r(r-p)
(b) Add: 2x(z – x – y) and 2y(z – y – x)
(c) Subtract: 3l (l – 4m + 5n) from 4l (10n – 3m + 2l)
(d) Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(-a + b + c)
Solution:
(a) āµ p(p -q) = p Ć p – p Ć q = p2 – pq
q(q – r) = q Ć q – q Ć r = q2 – qr
and r(r – p) = r Ć r – r Ć p = r2 – rp
ā“ Adding the above products, we have
(p2 – pq) + (q2 – qr) + (r2 – rp)
= p2 – pq + q2 – qr + r2 – rp
= p2 + q2 + r2 – pq – qr – rp
= p2 + q2 + r2 – (pq + qr + rp)

(b) āµ 2x(z – x – y) = 2x Ć z – 2x Ć x – 2x Ć y
= 2xz – 2x2 – 2xy and 2y(z – y – x)
= 2y Ć z – 2y Ć y – 2y Ć x
= 2yz – 2y2 – 2yx

(c) āµ 3l(l – 4m + 5n)
= 3l Ć l – 3l Ć 4m + 3l Ć 5n
= 3l2 – 12lm + 15ln and 4l(10n – 3m + 2l)
= 4l Ć 10n – 4l Ć 3m + 4l Ć 2l
= 40ln – 12lm + 8l2

(d) āµ 3a(a + b + c) – 2b(a – b + c)
= [3a Ć a + 3a Ć b + 3a Ć b + 3a Ć c] – 2b(a Ć a + 2b Ć (-b) + 2b Ć c]
= [3a2 + 3ab + 3ac] – [2ab – 2b2 + 2abc]
= 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc
= 3a2 + ab + 3ac + 2b2 – 2bc and 4c(-a + b + c)
= 4c Ć (-a) + 4c Ć b + 4c Ć c
= – 4ac + 4bc + 4c2