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

**Ex 12.3 Class 7 Maths Question 1.**

**If m = 2, find the value of:**

**(i) m – 2**

**(ii) 3m – 5**

**(iii) 9 – 5m**

**(iv) 3m2 – 2m – 7**

**(v) 5m/2 – 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) 5m/2 – 4

Putting m = 2, we get

5 × 2/2 – 4 = 5 – 4 = 1

**Ex 12.3 Class 7 Maths 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

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

(ii) -3p² + 4p + l

Putting p = -2, we get

-3(-2)² + 4(-2) + 7

= -3 × 4 – 8 + 7 = -12 – 8+ 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 = 3

**Ex 12.3 Class 7 Maths 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

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

(2)² + 2(-2) + (-2)² = 4 – 4 + 4 = 4

(iii) a² – b²

Putting a = 2 and b = -2, we get

(2)² – (-2)² = 4 – 4 = 0

**Ex 12.3 Class 7 Maths 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 which is required.

(ii) 2a² + b² + 1

= 2(0)² + (-1)² + 1 =0 + 1 + 1 = 2 which is required.

(iii) 2a²b + 2ab² + ab

= 2(0)² (-1) + 2(0)(-1)² + (0)(-1)

=0 + 0 + 0 = 0 which is required.

(iv) a² + ab + 2

= (0)² + (0)(-1) + 2

= 0 + 0 + 2 = 0 which is required.

**Ex 12.3 Class 7 Maths 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 = 5x – 13

Putting x = 2, we get

= 5 × 2 – 13 = 10 – 13 = -3

which is required.

(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1

Putting x = 2, we get

= 8 × 2 – 1 = 16 – 1 = 15

which is required.

(iii) 6x + 5(x – 2) = 6x + 5x – 10

= 11 × – 10

Putting x = 2, we get

= 11 × 2 – 10 = 22 – 10 = 12

which is required.

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11

= 11x + 7

Putting x = 2, we get

= 11 × 2 + 7 = 22+ 7 = 29

**Ex 12.3 Class 7 Maths 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 – 55**

**(v) 2a – 2b – 4 – 5 + a**

**Solution:**

(i) 3x – 5 – x + 9 = 2x + 4

Putting x = 3, we get

2 × 3 + 4 = 6 + 4 = 10

which is required.(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1

Putting x = 2, we get

= 8 × 2 – 1 = 16 – 1 = 15

which is required.

(iii) 6x + 5(x – 2) = 6x + 5x – 10

= 11 × – 10

Putting x = 2, we get

= 11 × 2 – 10 = 22 – 10 = 12

which is required.

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11

= 11x + 7

Putting x = 2, we get

= 11 × 2 + 7 = 22+ 7 = 29

**Ex 12.3 Class 7 Maths Question 8.**

**(i) If z = 10, find the value of z² – 3(z – 10).**

**(ii) If p = -10, find the value of p² -2p – 100.**

**Solution:**

(i) z² – 3(z – 10)

= z² – 3z + 30

Putting z = 10, we get

= (10)2 – 3(10) + 30

= 1000 – 30 + 30 = 1000 which is required.

(ii) p² – 2p – 100

Putting p = -10, we get

(-10)2 – 2(-10) – 100

= 100 + 20 – 100 = 20 which is required.

**Ex 12.3 Class 7 Maths Question 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

Putting x = 0, we get

2(0)² + (0) – a = 5

0 + 0 – a = 5

-a = 5

⇒ a = -5 which is required value.

**Ex 12.3 Class 7 Maths 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 + 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

Hence, the required value = 38.