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.
HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3 1
Solution:
(i) ab + ac, + ad
(ii) 5x2y + 5xy2 – 25xy
(iii) 6p3 – 7p2 + 5p
(iv) 4p4q2 – 4p2q4
(v) a2bc + ab2c + abc2.

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

Question 3.
Find the product.
(i) (a2) × (2a22) × (4a26)
HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3 2
(iv) x × x2 × x3 × x4.
Solution:
(i) a2 × 2 × a22 × 4 × a26 = 8a50.
HBSE 8th Class Maths Solutions Chapter 9 Algebraic Expressions and Identities Ex 9.3 3
(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.

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

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.

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