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

Question 1.

If m = 2, find the value of:

(i) m – 2

(ii) 3m – 5

(iii) 9 – 5m

(iv) 3m^{2} – 2m – 7

(v) \(\frac{5 m}{2}\) – 4

Solution:

(i) m – 2 = 2 – 2 = 0

(ii) 3m – 5 = (3 × 2) – 5 = 6 – 5 = 1

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

(iv) 3m^{2} – 2m – 7 = 3 × (2 × 2) – (2 × 2) – 7

= 12 – 4 – 7 = 1

(v) \(\frac{5 m}{2}\) – 4 = \(\left(\frac{5 \times 2}{2}\right)\) – 4 = 5 – 4 = 1

Question 2.

If p = -2, find the value of:

(i) 4p + 7

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

(iii) -2p^{3} – 3p^{2} + 4p + 7

Solution:

(i) 4p + 7 = 4 × (-2) + 7 = -8 + 7 = -1

(ii) -3p^{2} + 4p + 7 = -3 × (-2) × (-2) + 4 × (-2) + 7 = -12 – 8 + 7 = -13

(iii) -2p^{3} – 3p^{2} + 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) x^{2} + 2x + 1

(iv) 2x^{2} – x – 2

Solution:

(i) 2x – 7 = 2 × (-1) – 7 = -2 – 7 = -9

(ii) – x + 2 = – (-1) + 2 = 1 + 2 = 3

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

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

Question 4.

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

(i) a^{2} + b^{2}

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

(iii) a^{2} – b^{2}

Solution:

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

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

(iii) 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

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

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

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

Solution:

(i) 2a + 2b = 2 × (0) + 2 × (-1) = 0 – 2 = – 2

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

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

= 0 + 0 + 0 = 0

(iv) a^{2} + 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 z^{3} – 3(z – 10).

(ii) If p = -10, find the value of p^{2} – 2p -100.

Solution:

(i) For z = 10,

z^{3} – 3z + 30

= (10 × 10 × 10) – (3 × 10) + 30

= 1000 – 30 + 30 = 1000

(ii) For p = -10,

p^{2} – 2p – 100

=(-10) × (-10) – 2 × (-10) – 100

= 100 + 20 – 100 = 20

Question 9.

What should be the value of a if the value of 2x^{2} + x – a equals to 5, when x = 0?

Solution:

When x = 0; 2x^{2} + 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(a^{2} + ab) + 3 – ab

Solution:

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

= 2a^{2} + (2 – 1) ab + 3 = 2a^{2} + ab + 3

= 2 × (5 × 5) + 5 × (-3) + 3 [ ∵ a = 5, b = – 3]

= 50 – 15 + 3 = 38