PSEB 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.3

1. Carry out the multiplication of the expressions in each of the following pairs:

Question (i)
4p, q + r
Solution:
= 4p × (q + r)
= (4p × q) + (4p × r)
= 4pq + 4pr

Question (ii)
ab, a – b
Solution:
= ab × (a-b)
= (ab × a) – (ab × b)
= a²b – ab²

Question (iii)
a + b, 7a²b²
Solution:
= (a + b) × 7a²b²
= (a × 7a²b²) + (b × 7a²b²)
= 7 a³b² + 7a²b³

Question (iv)
a² – 9, 4a
Solution:
= (a² – 9) × 4a
= (a² × 4a) – (9 × 4a)
= 4a³ – 36a

Question (v)
pq + qr + rp, 0
Solution:
= (pq + qr + rp) × 0
= 0

2. Complete the table:

First expression	Second expression	Product
1. a	b + c + d	……………
2. x + y- 5	5xy	……………
3. P	6p2 – 7p + 5	…………..
4. 4p2q- q2	p2 – q2	…………..
5. a + b + c	abc	………….

Solution:
(i) a × (b + c + d)
= (a × b) + (a × c) + (a × d)
= ab + ac + ad

(ii) (x + y – 5) × 5xy
= (x × 5xy) + (y × 5xy) + [(- 5) × 5xy]
= 5x²y + 5xy² – 25xy

(iii) p × (6p² – 7p + 5)
= (p × 6p²) + [p × (- 7p)] + (p × 5)
= 6p³ – 7p² + 5p

(iv) 4p²q² × (p² – q²)
= (4p²q² × p²) + [4p²q² × (-q²)]
= 4p⁴q² – 4p²q⁴

(v) (a + b + c) × abc
= (a × abc) + (b × abc) + (c × abc)
= a²bc + ab²c + abc²

3. Find the product:

Question (i)
(a²) × (2a²²) × (4a²⁶)
Solution:
= (1 × 2 × 4) × a² × a²² × a²⁶
= 8 × a⁵⁰
= 8a⁵⁰

Question (ii)
(\(\frac {2}{3}\)xy) × (\(\frac {-9}{10}\)x²y²)
Solution:
= \(\frac {2}{3}\) × (\(\frac {-9}{10}\)) × xy × x²y²
= \(\frac {-2}{3}\) × \(\frac {9}{10}\) × x³y³
= \(\frac {-3}{5}\)x³y³

Question (iii)
(\(\frac {-10}{3}\)pq³) × (\(\frac {6}{5}\)p³q)
Solution:
= [(\(\frac {-10}{3}\)) × \(\frac {6}{5}\)] × pq³ × p³q
= – \(\frac {10}{3}\) × \(\frac {6}{5}\) × p⁴q⁴
= – 4p⁴q⁴

Question (iv)
x × x² × x³ × x⁴
Solution:
= (1 × 1 × 1 × 1) × x × x² × x³ × x⁴
= (1) × x¹⁰
= x¹⁰

4.

Question (a)
Simplify 3x(4x – 5) + 3 and find its value for
(i) x = 3
(ii) x = \(\frac {1}{2}\)
Solution:
3x (4x- 5) + 3
= (3x × 4x) – (3x × 5) + 3
= 12x¹⁰ – 15x + 3

(i) When x = 3, then
12x² – 15x + 3
= 12 (3)² – 15(3) + 3
= 12 (9) – 15 (3) + 3
= 108 -45 + 3
= 111 – 45
= 66

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

Question (b)
Simplify a (a2 + a + 1) + 5 and find its values for
(i) a = 0 (ii) a = 1 (iii) a = (-1)
Solution:
a (a² + a + 1) + 5
= (a × a²) + (a × a) + (a × 1) + 5
= a³ + a² + a + 5

(i) When a = 0, then
a³ + a² + a + 5
= (-1)³ + (0)³ + (0) + 5
= 0 + 0 + 0 + 5
= 5

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

(iii) a = (-1), then
a³ + a² + a + 5
= (-1)³ + (-1)² + (-1) + 5
= (-1) + (1) + (-1) + 5
= 6 – 2 = 4

5.

Question (a)
Add : p(p – q), q(q – r) and r(r – p)
Solution:
[p(p-q)] + [q(q-r)] + [r(r-p)]
= p² – pq + q² – qr + r² – rp
= p² + q² + r² – pq – qr – rp

Question (b)
Add : 2x(z – x – y) and 2y(z – y – x)
Solution:
[2x (z – x – y)] + [2y (z – y – x)]
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= – 2x² – 2y² – 2xy – 2xy + 2yz + 2xz
= – 2x² – 2y² – 4xy + 2yz + 2xz

Question (c)
Subtract : 31(l – 4m + 5n) from 4l(10n – 3m + 2l)
Solution:
[4l(10n – 3m + 21)] – [3l(l – 4m + 5n)]
= [40ln – 12lm + 8l²] – [3l² – 12lm + 15ln]
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 40ln – 15ln – 12lm + 12lm + 8l² – 3l²
= 25ln + 0lm + 5l²
= 25ln + 5l²

Question (d)
Subtract : 3a(a + b + c) – 2b(a – b + c) from 4c(- a + b + c)
Solution:
[4c (- a + b + c)] – [3a (a + b + c) – 2b (a – b + c)]
= [- 4ac + 4bc + 4c²]
– [3a² + 3ab + 3ac – 2ab + 2b² – 2bc]
= – 4ac + 4be + 4c² – 3a² – 3ab – 3ac + 2ab – 2b² + 2bc
= – 3a² – 2b² + 4c² – 3ab + 2ab + 4bc + 2bc – 4ac – 3ac
= – 3a² – 2b² + 4c² – ab + 6bc – 7ac