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

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

Haryana Board 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Exercise 9.3

Question 1.
Carry out the multiplication of the expressions in each of the following pairs :
(i) 4p, q + r
(ii) ab, a – b
(iii) a + b, 7a²b²
(iv) a² – 9, 4a
(v) pq + qr + rp, 0.
Solution:
(i) 4p(q + r) = 4pq + 4pr.
(ii) ab × (a – b) = a²b – ab².
(iii) (a + b) × (7a²b²) = 7a²b² + 7a²b³.
(iv) (a² – 9) × 4a = 4a³ – 36a.
(v) (pq – qr + rp) × 0 = 0

Question 2.
Complete the table.

Solution:
(i) ab + ac, + ad
(ii) 5x²y + 5xy² – 25xy
(iii) 6p³ – 7p² + 5p
(iv) 4p⁴q² – 4p²q⁴
(v) a²bc + ab²c + abc².

Question 3.
Find the product.
(i) (a²) × (2a²²) × (4a²⁶)

(iv) x × x² × x³ × x⁴.
Solution:
(i) a² × 2 × a²² × 4 × a²⁶ = 8a⁵⁰.

(iv) x × x² × x³ × x⁴ = x¹⁰.

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

(b) Simplify a (a² + a + 1) + 5 and find its value for
(i) a = 0
(ii) a = 1
(iii) a = -1.
Solution:
(a) 3x (4x – 5) + 3 = 12x² – 15x + 3
(i) If x = 3, 12 × (3)² – 15 × 3 + 3
= 12 × 9 – 45 + 3
= 108 – 42 = 66.

(ii) If x = \(\frac{1}{2}\) 12 × (\(\frac{1}{2}\))² – 15 × \(\frac{1}{2}\) + 3
= 12 × \(\frac{1}{4}\) – \(\frac{15}{2}\) + 3
= 6 – \(\frac{15}{2}\) = \(\frac{12-15}{2}\) = \(\frac{-3}{2}\)

(b) a(a² + a + 1) + 5 = a³ + a² + a + 5
(i) If a = 0
a³ + a² + a + 5 = 0³ + 0² + 0 + 5 = 5.

(ii) If a = 1,
a³ + a² + a + 5 = (1)³ + 1² + 1 + 5
= 1 + 1 + 1 + 5 = 8.

(iii) If a = 1
a³ + a² + a + 5 = (-1)³ + (-1)² + (-1) + 5
= 1 + 1 – 1 + 5 = 6.

Question 5.
(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) Add : (p² – pq) + (q² – qr) + (r² – rp)
= p² + q² + r² – pq – qr – rp.

(b) Add : (2xz – 2x² – 2xy) + (2yz – 2y² – 2xy)
= 2xz – 2x² – 2xy + 2yz – 2y² – 2xy
= -2x² – 2y² – 4xy + 2xz + 2yz.

(c) Subtract : (40ln – 12lm + 8l²) – (3l² – 12lm + 15ln)
= 40ln – 12lm + 8l² – 3l² + 12lm – 15ln
= 5l² + 25ln.

(d) Subtract : (-4ca + 4bc + 4c²) – (3a² + 3ab + 3ac – 2ab + 2b² – 2bc)
= (-4ca + 4bc + 4c²) – (3a² + 2b² + ab — 2bc + 3ac)
= -4ac + 45c + 4c² – 3a² – 2b² – ab + 25c – 3ac
= -3a² – 2b² + 4c² – ab + 6bc – 7ac.