MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

MP Board Class 7th Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:

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 = 4 × (-2) + 7 = -8 + 7 = -1
(ii) -3p² + 4p + 7 = -3 × (-2) × (-2) + 4 × (-2) + 7 = -12 – 8 + 7 = -13
(iii) -2p³ – 3p² + 4p + 7
= -2 × (-2) × (-2) × (-2) – 3 × (-2) × (-2) + 4 × (-2) + 7
= 16 – 12 – 8 + 7 = 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 = 2 × (-1) – 7 = -2 – 7 = -9
(ii) – x + 2 = – (-1) + 2 = 1 + 2 = 3
(iii) x² + 2x + 1 = (-1) × (-1) + 2 × (-1) + 1 = 1 – 2 + 1 = 0
(iv) 2x² – x – 2 = 2 × (-1) × (-1) – (-1) – 2 = 2 + 1 – 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 = 2 × 2 + (-2) × (-2) = 4 + 4 = 8
(ii) a² + ab + b² = (2 × 2) + 2 × (-2) + (-2) × (-2) = 4 – 4 + 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
(iii) 2a²b + 2ab² + ab
(iv) a² + ab + 2
Solution:
(i) 2a + 2b = 2 × (0) + 2 × (-1) = 0 – 2 = – 2
(ii) 2a² + b² + 1 = 2 × (0) × (0) + (-1) × (-1) + 1 = 0 + 1 + 1 = 2
(iii) 2a²b + 2ab² + ab = 2 × (0) × (0) × (-1) + 2 × (0) × (-1) × (-1) + 0 × (-1)
= 0 + 0 + 0 = 0
(iv) a² + ab + 2 = (0) × (0) + 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) = x + 7 + 4x – 20
= (1 + 4)x + (7 – 20) = 5x – 13
Putting x = 2 we get,
5x – 13 = (5 × 2) – 13 = 10 – 13 = -3

(ii) 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7
= (3 + 5)x + (6 – 7) = 8x – 1
Putting x = 2 we get,
8x – 1 = (8 × 2) – 1 = 16 – 1 = 15

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= (6 + 5)x – 10 = (11x – 10)
Putting x = 2 we get,
11x – 10 = (11 × 2) – 10 = 22 – 10 = 12

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11
= (8 + 3) × + (11 – 4) = 11x + 7
Putting x = 2 we get,
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
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8o + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a
Solution:
(i) 3x – 5 – x + 9 = (3 – 1) x + (-5 + 9)
= 2x + 4 = (2 × 3) + 4 [∵ x = 3]
= 6 + 4 = 10

(ii) 2 – 8x + 4x + 4 = 2 + 4 + (- 8 + 4)x
= 6 – 4x = 6 – (4 × 3) = 6 – 12 = -6 [∵ x = 3]

(iii) 3a + 5 – 8a +1 = (3 – 8)a + (5 + 1)
= -5a + 6
= -5 × (-1) + 6     [∵ a = -1]
= 5 + 6 = 11

(iv) 10 – 3b – 4 – 5b = 10 – 4 + (- 3 – 5)b
= 6 – 8b = 6 – 8 × (-2) [∵ b = -2]
= 6 + 16 = 22

(v) 2a – 2b – 4 – 5 + a = (2 + 1)a – 2b – 4 – 5
= 3a – 2b – 9
= 3 × (-1) – 2 × (-2) – 9 [∵ a = -1, b = -2]
= -3 + 4 – 9 = -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) For z = 10,
z³ – 3z + 30
= (10 × 10 × 10) – (3 × 10) + 30
= 1000 – 30 + 30 = 1000

(ii) For p = -10,
p² – 2p – 100
=(-10) × (-10) – 2 × (-10) – 100
= 100 + 20 – 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:
When x = 0; 2x² + x – a = 5,
∴ (2 × 0) + 0 – a = 5
⇒ 0 – 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:
2(a² + ab) + 3 – ab = 2a² + 2ab + 3 – ab
= 2a² + (2 – 1) ab + 3 = 2a² + ab + 3
= 2 × (5 × 5) + 5 × (-3) + 3       [ ∵ a = 5, b = – 3]
= 50 – 15 + 3 = 38

MP Board Class 7th Maths Solutions