CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3 Textbook Questions and Answers are provided by experts in order to help students secure good marks in exams.

## 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) 3m^{2} – 2m – 7

Solution:

3m^{2} – 2m – 7 = 3(2)^{2} – 2(2) – 7 = 3(4) – 4 – 7

(v) \(\frac{5 m}{2}\) – 4

Solution:

\(\frac{5 m}{2}\) – 4 = \(\frac{5 \times 2}{2}\) – 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) -3p^{2} + 4p + 7

Solution:

– 3p^{2} + 4p + 7 = – 3(- 2)^{2} + 4 (- 2) + 7

= -3(4) – 8 + 7

= -12 – 1

= -13

(iii) – 2p^{3} – 3p^{2} + 4p + 7

Solution:

– 2p^{3} – 3p^{2} + 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) x^{2} + 2x + 1

Solution:

x^{2} + 2x + 1 = (- 1)^{2} + 2(-1) + 1 = 1 – 2 + 1 = 2 – 2 = 0

(iv) 2x^{2} – x – 2

Solution:

2x^{2} – 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) a^{2} + b^{2}

Answer:

a^{2} + b^{2} = (2)^{2} + (- 2)^{2} = 4 + 4 = 8

(ii) a^{2} + ab + b^{2}

Solution:

a^{2} + ab + b^{2} = (2)^{2} + (2) (- 2) + (- 2) 2 = 4 – 4 + 4 = 4

(iii) a^{2} – b^{2}

Solution:

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

Solution:

2a + 2b = 0 + 2 (- 1) = – 2

(ii) 2a^{2} + b^{2} + 1

Solution:

2a^{2} + b^{2} + 1 = 0 + (-1)^{2} + 1 = 1 + 1 = 2

(iii) 2a^{2 }b + 2ab^{2 }+ ab

Solution:

2a^{2 }b + 2ab^{2} + ab = 0 + 0 + 0 = 0

(iv) a^{2} + ab + 2

Solution:

a^{2} + 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; z^{3} – 3(z – 10).

Solution:

When z = 10, then

z^{2} – 3(z -10) = (10)^{3} – 3 (10 -10)

= 1000 1 – 3(0)

= 1000 – 0

= 1000

(b) If p = – 10, find the value of p^{2} – 2p -10a

Solution:

When p = -10, then

p^{2} – 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= 0,2x^{2} + 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(a^{2} + ab) + 3 – ab

Solution:

When a = 5 and b = – 3, then,

2(a^{2} + ab) + 3 – ab = 2a^{2} + 2ab + 3 – ab

= 2a^{2} + ab + 3

At a = 5, b = – 3

2a^{2} + ab + 3 = 2(5)^{2} + (5) ( – 3) + 3

= 2 × 25 – 15 + 3

= 50 – 15 + 3 = 38