Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 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 – 5 m = 9 – 5(2) = 9 – 10 = – 1.

(iv) 3m^{2} – 2m – 7

= 3(2)^{2} – 2(2) – 7

= 3 × 4 – 4 – 7

= 12 – 4 – 7 = 8 – 7 = 1

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

= \(\frac{10}{2}\) – 4 = 5 – 4 = 1

Question 2.

If P = – 2, find the value of;

(i) 4p + 7

(ii) – 3p^{2} + 4p + 7

(iii) – 2p^{2} – 3p^{2} + 4p + 7, when p = – 2.

Solution:

(i) 4p + 7 = 4(- 2) + 7 = – 8 + 7

= -1

(ii) – 3(-p)^{2} + 4(-p) + 7

= – 3(- 2)^{2} + 4(- 2) + 7

= – 3 × 4 + 4 × -2 + 7

= – 12 – 8 + 7

= – 13.

(iii)- 2p^{2} – 3p^{2} + 4p + 7

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

= -2 × 4 – 3 × 4 + 4 × – 2 + 7

= -8 – 12 – 8 + 7

= -21.

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)^{2} + 2(- 1) + 1

= 1 – 2 + 1 = 0.

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

= 2 × 1 + 1 – 2

= 2 + 1 – 2

= 3 – 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} – b2^{2}.

Solution:

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

= 4 + 4

= 8

(ii) a^{2} + ab + b^{2} = (2)^{2} + 2 x -2 + (-2)^{2}

= 4 – 4 + 4

= 8 – 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

(Hi) 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)^{2} + (- 1)^{2} + 1

= 2 × 0 + 1 + 1

= 1 + 1 = 2.

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

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

= 2 × 0 × (-1) + 2 × 0 × 1 + 0

= 0 + 0 + 0 = 0

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

a = 0

b = -1

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

= 0 + 0 + 2 = 2.

Question 6.

Simplify the expressions and find the value of x is 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

= 2 + 7 + 4 × 2 – 20

= 9 + 8 – 20

= 17 – 20 = – 3.

(ii) 3(x + 2) + 5x – 7

= 3x + 6 + 5x – 7

= 3 × 2 + 6 + 5 × 2 – 7

= 6 + 6 + 10 – 7

= 12 + 10 – 7

= 22 – 7 = 15.

(iii) 6x + 5(x – 2) = 6x + 5x – 10

= 6 × 2 + 5 × 2 – 10

= 12 + 10 – 10

= 22 – 10 = 12.

(iv) 4(2x – 1) + 3x + 11

= 8x – 4 + 3x + 11

= 8 x 2 – 4 + 3 x 2 + 11

= 16 – 4 + 6 + 11

= 12 + 6 + 11

= 18 + 11

= 29.

Question 7.

Simplify these expressions and find their yalue of 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 – 5b

(v) 2a – 2b – 4 – 5 + a.

Solution:

(i)3x – 5 – x + 9

= 3 × 3 – 5 – 3 + 9

= 9 – 5 – 3 + 9

= 4 – 3 + 9

= 1 + 9 = 10.

(ii) 2 – 8x + 4x + 4

= 2 – 8 × 3 + 4 × 3 + 4

= 2 – 24 + 12 + 4

= 18 – 24 = – 6.

(iii) 3a + 5 – 8a + 1

= 3(- 1) + 5 – 8(- 1) + 1

= -3 + 5 + 8 + 1

= 2 + 8 + 1 = 11.

(iv) 10 – 3b – 4 – 5b

= 10 – 3(-2) – 4 – 5(-2)

= 10 + 6 – 4 + 10

= 16 – 4 + 10 = 12 + 10

= 22.

(v) 2a -26-4-5 + a

= 2(-1) – 2(-2) – 4 – 5 + (-1)

= -2 + 4 – 4 – 5 – 1

= 2 – 9 – 1

= 2 – 10 = -8.

Question 8.

(a) If z = 10, find the value of z^{3} – 3(z -10).

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

Solution:

(a) z^{3} – 3(z -10)

= (10)^{3} – 3(10-10)

= 1000 – 3 x 0

= 1000 – 0 = 1000

(b) p^{2} – 2p – 100 = (- 10)^{2} – 2 x (- 10) – 100

= 100 + 20 – 100

= 120 – 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:

2x^{2} + x – a = 5

2(0)^{2} + 0 – a = 5

or 2 x 0 + 0 – a = 5

or -a = 5 or 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} + ab + 3

= 2 × (5)^{2} + (5) × (-3) + 3

= 2 × 25 + (- 15) + 3

= 50 – 15 + 3

= 53 – 15 = 38.