NCERT Solutions Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

NCERT Solutions Class 7 Mathematics 
Chapter – 12 (Algebraic Expressions)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 12 Algebraic Expressions  Exercise 12.3 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 12 Algebraic Expressions

• NCERT Solution Class 7 Maths Exercise – 12.1
• NCERT Solution Class 7 Maths Exercise – 12.2
• NCERT Solution Class 7 Maths Exercise – 12.4

Exercise – 12.3 

1. If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) − 4

Solution –

(i) m – 2
Putting m = 2, we get
2 – 2
= 0

(ii) 3m – 5
Putting m = 2, we get
3 × 2 – 5
= 6 – 5
= 1

(iii) 9 – 5m
Putting m = 2, we get
9 – 5 × 2
= 9 – 10
= -1

(iv) 3m² – 2m – 7
Putting m = 2, we get
3(2)² – 2(2) – 7
= 3 × 4 – 4 – 7
= 12 – 4 – 7
= 12 – 11
= 1

(v) − 4
Putting m = 2, we get
– 4
= 5 − 4
= 1

2. If p = – 2, find the value of:
(i) 4p + 7
(ii) -3p² + 4p + 7
(iii) -2p³ – 3p² + 4p + 7

Solution –

(i) 4p + 7
Putting p = -2, we get
= (4 × (-2)) + 7
= -8 + 7
= -1

(ii) – 3p² + 4p + 7
Putting p = -2, we get
= (-3 × (-2)²) + (4 × (-2)) + 7
= (-3 × 4) + (-8) + 7
= -12 – 8 + 7
= -20 + 7
= -13

(iii) – 2p³ – 3p² + 4p + 7
Putting p = -2, we get
= (-2 × (-2)³) – (3 × (-2)²) + (4 × (-2)) + 7
= (-2 × -8) – (3 × 4) + (-8) + 7
= 16 – 12 – 8 + 7
= 23 – 20
= 3

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
Putting x = -1, we get
= (2 × -1) – 7
= – 2 – 7
= – 9

(ii) – x + 2
Putting x = -1, we get
= – (-1) + 2
= 1 + 2
= 3

(iii) x² + 2x + 1
Putting x = -1, we get
= (-1)² + (2 × -1) + 1
= 1 – 2 + 1
= 2 – 2
= 0

(iv) 2x² – x – 2
Putting x = -1, we get
= (2 × (-1)²) – (-1) – 2
= (2 × 1) + 1 – 2
= 2 + 1 – 2
= 3 – 2
= 1

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

Solution –

(i) a² + b²
Putting a = 2 and b = -2, we get
= (2)² + (-2)²
= 4 + 4
= 8

(ii) a² + ab + b²
Putting a = 2 and b = -2, we get
= 2² + (2 × -2) + (-2)²
= 4 + (-4) + (4)
= 4 – 4 + 4
= 4

(iii) a² – b²
Putting a = 2 and b = -2, we get
= 2² – (-2)²
= 4 – (4)
= 4 – 4
= 0

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
Putting a = 0 and b = -1, we get
= (2 × 0) + (2 × -1)
= 0 – 2
= -2

(ii) 2a² + b² + 1
Putting a = 0 and b = -1, we get
= (2 × 0²) + (-1)² + 1
= 0 + 1 + 1
= 2

(iii) 2a²b + 2ab² + ab
Putting a = 0 and b = -1, we get
= (2 × 0² × -1) + (2 × 0 × (-1)²) + (0 × -1)
= 0 + 0 +0
= 0

(iv) a² + ab + 2
Putting a = 0 and b = -1, we get
= (0²) + (0 × (-1)) + 2
= 0 + 0 + 2
= 2

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
= 5x + 7 – 20
Putting x = 2, we get
= (5 × 2) + 7 – 20
= 10 + 7 – 20
= 17 – 20
= – 3

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

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

(iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 11x + 7
Putting x = 2, we get
= (11 × 2) + 7
= 22 + 7
= 29

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 – 55
(v) 2a – 2b – 4 – 5 + a

Solution –

(i) 3x – 5 – x + 9
= 3x – x – 5 + 9
= 2x + 4
Putting x = 3, we get
= (2 × 3) + 4
= 6 + 4
= 10

(ii) 2 – 8x + 4x + 4
= 2 + 4 – 8x + 4x
= 6 – 4x
Putting x = 3, we get
= 6 – (4 × 3)
= 6 – 12
= – 6

(iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= – 5a + 6
Putting a = -1, we get
= – (5 × (-1)) + 6
= – (-5) + 6
= 5 + 6
= 11

(iv) 10 – 3b – 4 – 5b
= 10 – 4 – 3b – 5b
= 6 – 8b
Putting b = -2, we get
= 6 – (8 × (-2))
= 6 – (-16)
= 6 + 16
= 22

(v) 2a – 2b – 4 – 5 + a
= 2a + a – 2b – 4 – 5
= 3a – 2b – 9
Putting a = -1 and b = -2, we get
= (3 × (-1)) – (2 × (-2)) – 9
= -3 – (-4) – 9
= – 3 + 4 – 9
= -12 + 4
= -8

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) If z = 10, find the value of z² – 3(z – 10).
= z² – 3(z – 10)
= z² – 3z + 30
Putting z = 10, we get
= (10)² – 3(10) + 30
= 1000 – 30 + 30
= 1000

(ii) If p = – 10, find the value of p² – 2p – 100
Putting p = -10, we get
= (-10)² – 2(-10) – 100
= 100 + 20 – 100
= 20

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
a = 2x² + x – 5
Putting x = 0, we get
a = (2 × 0²) + 0 – 5
a = 0 + 0 – 5
a = -5

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 + 2ab – ab + 3
= 2ab + ab + 3
Putting, a = 5 and b = -3, we get
= 2(5)² + (5)(-3) + 3
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 53 – 15
= 38