# HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.3

Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m2 – 2m – 7
(v) $$\frac{5 m}{2}$$ – 4
Solution:
(i) m – 2 – 2 – 2 = 0
(ii) 3m – 5 = 3(2) – 5 = 6 – 5 = 1
(iii) 9 – 5 m = 9 – 5(2) = 9 – 10 = – 1.
(iv) 3m2 – 2m – 7
= 3(2)2 – 2(2) – 7
= 3 × 4 – 4 – 7
= 12 – 4 – 7 = 8 – 7 = 1
(v) $$\frac{5 m}{2}$$ – 4 = $$\frac{5 \times 2}{2}$$ – 4
= $$\frac{10}{2}$$ – 4 = 5 – 4 = 1

Question 2.
If P = – 2, find the value of;
(i) 4p + 7
(ii) – 3p2 + 4p + 7
(iii) – 2p2 – 3p2 + 4p + 7, when p = – 2.
Solution:
(i) 4p + 7 = 4(- 2) + 7 = – 8 + 7
= -1

(ii) – 3(-p)2 + 4(-p) + 7
= – 3(- 2)2 + 4(- 2) + 7
= – 3 × 4 + 4 × -2 + 7
= – 12 – 8 + 7
= – 13.

(iii)- 2p2 – 3p2 + 4p + 7
= -2(-2)2 – 3(- 2)2 + 4 × (- 2) + 7
= -2 × 4 – 3 × 4 + 4 × – 2 + 7
= -8 – 12 – 8 + 7
= -21.

Question 3.
Find the value of the following expressions, when x = – 1.
(i) 2x- 7
(ii) -x + 2
(iii) x2 + 2x + 1
(iv) 2x2 – x – 2.
Solution:
(i) 2x – 7 = 2(-1) – 7 = -2 – 7
= -9

(ii) -x + 2 = – (-1) + 2
= 1 + 2 = 3

(iii) x2 + 2x + 1
= (- 1)2 + 2(- 1) + 1
= 1 – 2 + 1 = 0.

(iv) 2x2 – x – 2 = 2(- 1)2 – (- 1) – 2
= 2 × 1 + 1 – 2
= 2 + 1 – 2
= 3 – 2 = 1.

Question 4.
If a = 2, b = – 2, find the value of:
(i) a2 + b2
(ii) a2 + ab + b2
(iii) a2 – b22.
Solution:
(i) a2 + b2 = (2)2 + (- 2)2
= 4 + 4
= 8

(ii) a2 + ab + b2 = (2)2 + 2 x -2 + (-2)2
= 4 – 4 + 4
= 8 – 4 = 4.

(iii) a2 – b2 = (2)2 – (- 2)2
= 4 – 4
= 0.

Question 5.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a+2b
(ii) 2a2 + b2 + 1
(Hi) 2a2b + 2ab2 + ab
(iv) a2 + ab + 2.
Solution:
(i) 2a + 2b = 2 × 0 + 2(- 1)
= 0 + (- 2) = – 2

(ii) 2a2 + b2 + 1 = 2(0)2 + (- 1)2 + 1
= 2 × 0 + 1 + 1
= 1 + 1 = 2.

(iii) 2a2b + 2ab2 + ab
= 2(0)2 (- 1) + 2 × 0(- 1)2 + 0 × – 1
= 2 × 0 × (-1) + 2 × 0 × 1 + 0
= 0 + 0 + 0 = 0

(iv) a2 + ab + 2
a = 0
b = -1
= (0)2 + (0) (- 1) + 2
= 0 + 0 + 2 = 2.

Question 6.
Simplify the expressions and find the value of x is 2.
(i) x + 7 + 4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x-2)
(iv) 4(2x – 1) + 3x + 11
Solution:
(i) x + 7 + 4(x – 5)
= x + 7 + 4x – 20
= 2 + 7 + 4 × 2 – 20
= 9 + 8 – 20
= 17 – 20 = – 3.

(ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x – 7
= 3 × 2 + 6 + 5 × 2 – 7
= 6 + 6 + 10 – 7
= 12 + 10 – 7
= 22 – 7 = 15.

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= 6 × 2 + 5 × 2 – 10
= 12 + 10 – 10
= 22 – 10 = 12.

(iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8 x 2 – 4 + 3 x 2 + 11
= 16 – 4 + 6 + 11
= 12 + 6 + 11
= 18 + 11
= 29.

Question 7.
Simplify these expressions and find their yalue of x = 3, a = – 1, b = – 2,
(i) 3x – 5 -x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a.
Solution:
(i)3x – 5 – x + 9
= 3 × 3 – 5 – 3 + 9
= 9 – 5 – 3 + 9
= 4 – 3 + 9
= 1 + 9 = 10.

(ii) 2 – 8x + 4x + 4
= 2 – 8 × 3 + 4 × 3 + 4
= 2 – 24 + 12 + 4
= 18 – 24 = – 6.

(iii) 3a + 5 – 8a + 1
= 3(- 1) + 5 – 8(- 1) + 1
= -3 + 5 + 8 + 1
= 2 + 8 + 1 = 11.

(iv) 10 – 3b – 4 – 5b
= 10 – 3(-2) – 4 – 5(-2)
= 10 + 6 – 4 + 10
= 16 – 4 + 10 = 12 + 10
= 22.

(v) 2a -26-4-5 + a
= 2(-1) – 2(-2) – 4 – 5 + (-1)
= -2 + 4 – 4 – 5 – 1
= 2 – 9 – 1
= 2 – 10 = -8.

Question 8.
(a) If z = 10, find the value of z3 – 3(z -10).
(6) If p = -10 find the value of p2 – 2p – 100.
Solution:
(a) z3 – 3(z -10)
= (10)3 – 3(10-10)
= 1000 – 3 x 0
= 1000 – 0 = 1000

(b) p2 – 2p – 100 = (- 10)2 – 2 x (- 10) – 100
= 100 + 20 – 100
= 120 – 100 = 20.

Question 9.
What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0 ?
Solution:
2x2 + x – a = 5
2(0)2 + 0 – a = 5
or 2 x 0 + 0 – a = 5
or -a = 5 or a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3, 2(a2 + ab) + 3 – ab.
Solution:
2(a2 + ab) + 3 – ab
= 2a2 + 2ab + 3 – ab
= 2a2 + ab + 3
= 2 × (5)2 + (5) × (-3) + 3
= 2 × 25 + (- 15) + 3
= 50 – 15 + 3
= 53 – 15 = 38.