# GSEB Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 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) $$\frac { 5m }{ 2 }$$ – 4
Solution:
(i) m – 2
∵ m = 2
∴ m – 2 = 2 – 2 = 0

(ii) 3m – 5
∵ m = 2
3(m) – 5 = 3(2) – 5
= 6 – 5 = 1

(iii) 9 – 5m
∵ m = 2
∴ 9 – 5(m) = 9 – 5(2)
= 9 – 10
= -1

(iv) 3m² – 2m – 7
∵ m = 2
∴ 3(m)² – 2(m) – 7 = 3(2)² – 2(2) – 7
= 3(4) – 4 – 7
= 12 – 11 = 1

(v) $$\frac { 5m }{ 2 }$$ – 4
∵ m = 2
∴ $$\frac { 5(m) }{ 2 }$$ – 4 = $$\frac { 5(2) }{ 2 }$$ – 4
= 5 – 4 = 1 Question 2.
If p = – 2 find the value of:
(i) 4p + 7
(ii) – 3p² + 4p + 7
(iii) – 2p3 – 3p² + 4p + 7
Solution:
(i) 4p + 7
∵ P = – 2
∴ 4(p) + 7 = 4(- 2) + 7
= – 8 + 7 = – 1

(ii) – 3p² + 4p + 7
∵ p = – 2
∴ – 3(p)² + 4(p) + 7 = – 3(- 2)² + 4(- 2) + 7
= – 3(4) + (- 8) + 7 = – 12 – 8 + 7 = – 20 + 7 = – 13

(iii) – 2p3 – 3p2 + 4p + 7
∵ P = -2
∴ – 2(p)3 – 3(p)2 + 4(P) + 7
= – 2(- 2)3 – 3(- 2)2 + 4(-2) + 7
= -2(- 8) – 3(4) + 4(-2) + 7
= 16 – 12 – 8 + 7
= 23 – 20 = 3

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
∵ x = – 1
∴ 2(x) – 7 = 2(-1) – 7
= – 2 – 1 = – 9

(ii) – x + 2
∵ x = – 1
∴ – (x) + 2 = -(- 1) + 2
= 1 + 2 = 3

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

(iv) 2x² – x – 2
∵ x = – 1
∴ 2(x)² – 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 – b²
Solution:
(i) a² + b²
∵ a = 2, and b = -2
∴ (a)² + (b)² =(2)² + (-2)² = 4 + 4 = 8

(ii) a² + ab +b²
∵ a = 2 and b = -2
(a)² + (a)(b) + (b)² = (2)² + (2)(-2) + (2)²
=4 – 4 + 4 = 8 – 4 = 4

(iii) a² – b²
∵ a = 2 and b = – 2
∴ (a)² – (b)² = (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
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Solution:
(i) 2a + 2b
∵ a = 0 and b = – 1
∴ 2(a) + 2 (b) = 2(0) + 2(-1)
= 0 – 2 = -2

(ii) 2a² + b² + 1
∵ a = 0 and b = – 1
∴ 2(a)² + (b)² + 1
= 2(0) + (-1)² + 1
= 2(0) + 1 + 1
= 0 + 2 = 2

(iii) 2a²b + 2ab² + ab
∵ a = 0 and b = -1
∴ 2(a)²(b) + 2(a)(b)² + (a)(b)
= 2(0)²(-1) + 2(0)(-1) + (0)(-1)
= 2(0)(-1) + 2(0)(1) + (0)(-1)
= 0 + 0 + 0 = 0

(iv) a² + ab + 2
∵ a = 0 and b = – 1
∴ (a)² + (a)(b) + 2 = (0)² + (0)(-1) + 2
= 0 + 0 + 2 = 2 Question 6.
Simplify the expressions and find the value if x is equal to 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)
We have: x + 7 + 4(x – 5) = x + 7 + 4x – 20
= 5x – 13
Now, 5(x) – 13 = 5(2) – 13 [∵ x = 2]
= 10 – 13 = – 3

(ii) 3(x + 2) + 5x – 7
We have: 3(x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= (3x + 5x) + (6 – 7)
= 8x – 1
Now, 8x – 1 = 8(x) – 1
= 8(2) – 1 (∵ x = 2)
= 16 – 1 = 15

(iii) 6x + 5(x – 2)
We have: 6x + 5(x – 2) = 6x + 5x – 10
= (6x + 5x) – 10
= 11x – 10
Now, 11x – 10= 11(x) – 10
= 11(2) – 10 [∵ x = 2]
= 22 – 10 = 12

(iv) 4(2x – 1) + 3x + 11
We have: 4 (2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= (8x + 3x) – 4 + 11
= 11x + 7
Now, 11x + 7= 11(x) + 7
= 11(2) + 7 [∵ x = 2]
= 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if 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
We have: 3x – 5 – x + 9
= (3x – x) – 5 + 9 = 2x + 4
Now, 2x + 4 = 2(x) + 4
= 2(3)+ 4 (∵ x = 3)
= 6 + 4 = 10

(ii) 2 – 8x + 4x + 4
We have: 2 – 8x + 4x + 4 = (-8x + 4x) + (2 + 4)
= – 4x + 6
Now, – 4x + 6 = – 4(x) + 6
= – 4(3) + 6 (∵ x = 3)
= – 12 + 6 = – 6

(iii) 3a + 5 – 8a + 1
We have: 3a + 5 – 8a + 1 = (3a – 8a) + (5 + 1)
= – 5a + 6
Now, – 5a + 6 = – 5(a) + 6
= – 5(-1) + 6 (∵ a = – 1)
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b
We have: 10 – 3b – 4 – 56
= (10 – 4) + (- 3b – 5b)
= 6 – 8b
Now, 6 – 8b = 6 – 8(b)
= 6 – 8(- 2) (∵ b = – 2)
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a
We have: 2a – 2b – 4 – 5 + a
= (2a + a) – 2b + (- 4 – 5)
= 3a – 2b + (- 9)
= 3a – 2b – 9
Now, 3a – 2b – 9 = 3(a) – 2(b) – 9 (∵ a = – 1, b = – 2)
= 3(- 1) – 2(- 2) – 9
= – 3 + 4 – 9
= – 12 + 4 = – 8

Question 8.
(i) If z = 10,find the value of z3 – 3(z – 10).
(ii) If p = – 10, find the value of p² – 2p – 100.
Solution:
(i) z³ – 3(z – 10) at z = 10
We have: z³ – 3(z – 10) = (z)³ – 3[(z) – 10]
= (10)³ – 3[10 – 10]
= 1000 – 3(0)
= 1000 – 0 = 1000

(ii) p² – 2p – 100 at p = – 10
We have: p² – 2p – 100 = (p)² – 2(p) – 100
= (- 10)² – 2(- 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:
Since, 2x² + x – a = 5 and x = 0
∴ 2(x)² + (x) – a = 5
or 2(0)² + (0) – a = 5
or 0 + 0 – a= 5
or – a = 5 ⇒ a = – 5
Thus, the required value of a = – 5.

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