NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Class 7 Maths NCERT Solutions Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
(i) m – 2
Solution :
m – 2 = 2 – 2 = 0

(ii) 3m – 5
Solution:
3m – 5 = 3 × 2 – 5 = 6 – 5 = 1

(iii) 9 – 5m
Solution:
9 – 5m = 9 – 5 × 2 = 9 – 10 = -1

(iv) 3m² – 2m  – 7
Solution:
3m² – 2m – 7 = 3(2)² – 2(2) – 7 = 3(4) – 4 – 7

(v) – 4
Solution:
– 4 = – 4
= 5 – 4
= 1

Question 2.
If p = -2, find the value of:

(i) -4p + 7
Solution:
4p + 7 = 4(- 2) + 7 = – 8 + 7 = – 1

(ii) -3p² + 4p + 7
Solution:
– 3p² + 4p + 7 = – 3(- 2)² + 4 (- 2) + 7
= -3(4) – 8 + 7
= -12 – 1
= -13

(iii) – 2p³ – 3p² + 4p + 7
Solution:
– 2p³ – 3p² + 4p + 7
= – 2(- 2)³ – 3(- 2)² + 4(- 2) + 7
= – 2(- 8) – 3(4) – 8 + 7
= 16 -12 – 8 + 7 = 23 – 20
= 3

Question 3.
Find the value of the following expressions, when x = -1

(i) 2x – 7
Solution:
2x – 7 = 2 (- 1) – 7 = – 2 – 7 = -9

(ii) – x + 2
Solution:
– x + 2 = – (-1) + 2 = 1 + 2 = 3

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

(iv) 2x² – x – 2
Solution:
2x² – x – 2 = 2( – 1)² – (- 1) – 2 = 2(1) + 1 – 2 = 2 + 1 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of:

(i) a² + b²
Answer:
a² + b² = (2)² + (- 2)² = 4 + 4 = 8

(ii) a² + ab + b²
Solution:
a² + ab + b² = (2)² + (2) (- 2) + (- 2) 2 = 4 – 4 + 4 = 4

(iii) a² – b²
Solution:
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
Solution:
2a + 2b = 0 + 2 (- 1) = – 2

(ii) 2a² + b² + 1
Solution:
2a² + b² + 1 = 0 + (-1)² + 1 = 1 + 1 = 2

(iii) 2a² b + 2ab² + ab
Solution:
2a² b + 2ab² + ab = 0 + 0 + 0 = 0

(iv) a² + ab + 2
Solution:
a² + ab + 2 = 0 + 0 + 2 = 2

Question 6.
Simplify the expressions and find the value of x is equal to 2 :

(i) x + 7 + 4 (x – 5)
Solution:
x + 7 + 4(x – 5) = x + 7 + 4x – 20
= (x + 4x) + (7 – 20)
= 5x – 13
At x = 2,
5x – 13 = 5(2) – 13
10 – 13 = -3

(ii) 3(x + 2) + 5x – 7
Solution:
3(x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= (3x + 5x) + (6 – 7)
= 8x -1
At x = 2,
8x – 1 = 8(2) – 1
= 16 – 1 = 15

(iii) 6x + 5(x – 2)
Solution:
6x + 5(x – 2) = 6 x + 5x – 10
= 11x – 10
At x = 2,
11x – 10 =11 × 2 – 10
= 22 – 10 = 12

(iv) 4(2x – 1) + 3x + 11
Solution:
4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= (8x +3x) + (- 4 +11)
= 11x + 7
At x = 2,
11x + 7 = 11 × 2 + 7 = 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = – 1, b = 2.

(i) 3x – 5 – x + 9
Answer:
3x – 5 – x + 9 = 2x + 4
At x = 3,
2x + 4 = 2(3) + 4
= 6 + 4 =10

(ii) 2 – 8x + 4x + 4
Solution :
2 – 8x + 4x + 4 = 6 – 4x
At x = 3,
6 – 4x = 6 – 4(3)
= 6 – 12 = -6

(iii) 3a + 5 – 8a + 1
Solution:
3a + 5 – 8a + 1 = – 5a + 6
At a = – 1,
– 5a + 6 = – 5 (- 1) + 6
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b
Solution:
10 – 3b – 4 – 5b = 6 – 8 b
At b = – 2,
6 – 8 b = 6 – 8(-2)
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a
Solution:
2a – 2b – 4 – 5 + a = 3a – 2b – 9
At a = -1, b = 2,
3a – 2b – 9 = 3(- 1) – 2(- 2) – 9
= -3 + 4 – 9
= -12 + 4
= -8

Question 8.
(a) If z = 10, find the value of; z³ – 3(z – 10).
Solution:
When z = 10, then
z² – 3(z -10) = (10)³  – 3 (10 -10)
= 1000 1 – 3(0)
= 1000 – 0
= 1000

(b) If p = – 10, find the value of p² – 2p -10a
Solution:
When p = -10, then
p² – 2p – 100 = (- 10)² – 2(- 10) – 100
= 100 + 20 – 100 = 20

Question 9.
What should be the value of an if the value of 2x to 5, when an r = 0?
Solution:
At x= 0,2x² + x – a = 5 (Given)
=> 2(0) + 0 – a = 5
=> 0 + 0 – a = 5
=> – a = 5
=> a = – 5

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