# HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.2 Textbook Exercise Questions and Answers.

## Haryana Board 7th Class Maths Solutions Chapter 4 Simple Equations Exercise 4.2

Question 1.
Give first the step you will use to separate the variable and then solve the equation:
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y -4 = – 7
(f) y -4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4.
Solution:
(a) x – 1 = 0
L.H.S. = x- R.H.S.
1 + 1 = x = 0 + 1 = 1
∴ x = 1
Which is the required solution.

(b) x + 1 = 0
L.H.S. = x + 1 – 1 = x
R.H.S. = 0 – 1 = -1
∴ x = -1
Which is the required solution.

(c) x – 1 = 5
L.H.S. = x – 1 + 1 = x
R.H.S. = 5 + 1 = 6
∴ x = 6
Which is the required solution.

(d) x + 6 =2
L.H.S. = x + 6 – 6 = x
R.H.S. = 2 – 6 = – 4
∴ x = -4
Which is the required solution.

(e) y- 4 = -7
L.H.S. = y – 4 + 4 = y
R.H.S. = – 7 + 4 = – 3
∴ y = -3
Which is the required solution.

(f) y – 4 = 4
L.H.S. = y – 4 + 4 = y
R.H.S. = 4 + 4 = 8
∴ y = 8
Which is the required solution.

(g) y + 4 = 4
L.H.S. = y + 4 + 4= y
R.H.S. = 4 – 4 = 0
∴ y = 0
Which is the required solution.

(h) y + 4 = – 4
L.H.S. = y + 4 — 4 = y
R.H.S. = – 4 – 4 = – S
∴ y = -8
Which is the required solution.

Question 2.
Give first the step you will use to separate the variable and then solve the equation.
(a) 3l = 42
(b) $$\frac{b}{2}$$ = 6
(c) $$\frac{p}{7}$$ = 4
(d) 4x = 25
(e) 8y = 36
(f) $$\frac{z}{3}=\frac{5}{4}$$
(g) $$\frac{a}{5}=\frac{7}{15}$$
(h) 20t = – 10.
Solution:
(a) 3l = 42
L.H.S. = 3l = $$\frac{3}{3}$$ x l = l
R.H.S. = 42 = $$\frac{42}{3}$$ = 14
∴ l = 14
Which is the required solution.

b) $$\frac{b}{2}$$ = 6
L.H.S = $$\frac{b}{2}=\frac{b}{2}$$ x 2 = b
R.H.S = 6 = 6 x 2 = 12
∴ b = 12.
Which is the required solution.

(c) $$\frac{p}{7}$$ = 4
L.H.S. = $$\frac{p}{7}=\frac{p}{7}$$ x 7 = p
R.H.S. = 4 = 4 x 7 = 28
∴ p = 28
Which is the required solution.

(d) 4 x = 25
L.H.S = 8y = $$\frac{8y}{8}$$ = y
R.H.S = 36 = $$\frac{36}{8}=\frac{9}{2}$$
∴ y = $$\frac{9}{2}$$
Which is the required solution.

(f) $$\frac{z}{3}=\frac{5}{4}$$
L.H.S = $$\frac{z}{3}=\frac{z}{3}$$ x 3 = z
R.H.S = $$\frac{5}{4}=\frac{5}{4} \times 3=\frac{15}{4}$$
∴ y = $$\frac{15}{4}$$
Which is the required solution.

(f) $$\frac{a}{5}=\frac{7}{15}$$
L.H.S. = $$\frac{a}{5}=\frac{a}{5}$$ x 5 = a
R.H.S = $$\frac{7}{15}=\frac{7}{15}$$ x 5 = $$\frac{7}{3}$$
∴ a = $$\frac{7}{3}$$
Which is the required solution.

(h) 20t = -10
L.H.S. = 20t = $$\frac{20 t}{20}$$ = t
R.H.S = -10 = $$\frac{-10}{20}=-\frac{1}{2}$$
∴ t = $$-\frac{1}{2}$$
Which is the required solution.

Question 3.
Give the step you will use to separate the variable and then solve the equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) $$\frac{20 p}{3}$$ = 40
(d) $$\frac{3 p}{10}$$ = 6
Solution:
3n – 2 = 46
L.H.S. = 3n – 2 + 2
= 3n = $$\frac{3n}{3}$$ = n
R.H.S. = 46 + 2 = 48 = $$\frac{48}{3}$$ = 16
∴ n = 16
Which is the required solution.

(b) 5m + 7 = 17
L.H.S. = 5 m + 7 = 5m + 7-7
= 5m = $$\frac{5m}{5}$$ = m
R.H.S = 17 = 17 – 7 = 10
= $$\frac{10}{5}$$ = 2
∴ m = 2
Which is the required solution.

(c) $$\frac{20 p}{3}$$ = 40
L.H.S = $$\frac{20 p}{3}=\frac{20 p}{3} \times 3$$
= 20 p = $$\frac{20 p}{20}$$ = p
R.H.S. = 40 = 40 x 3
= 120 = $$\frac{120}{20}$$ = 6
∴ P = 6
Which is the required solution.

(d) $$\frac{3 p}{10}$$ = 6
L.H.S.
= $$\frac{3 p}{10}=\frac{3 p}{10} \times 10=3 p=\frac{3 p}{3}=p$$
R.H.S. = 6 x 10 = 60 = $$\frac{60}{3}$$ = 20
∴ p = 20
Which is the required solution.

Question 4.
Solve the following equations
(a) 10p = 100
(b) 10p + 10 = 100
(c) $$\frac{p}{4}$$ = 5
(d) $$\frac{-p}{3}$$ = 5
(e) $$\frac{3 p}{4}$$ = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
Solution:
(a) 10p = 100
$$\frac{10 p}{10}=\frac{100}{10}$$
∴ p = 10

(b) 10p + 10 = 100
10p = 100 – 10
10p = 90
$$\frac{10 p}{10}=\frac{90}{10}$$
∴ p = 9

(c) $$\frac{p}{4}$$ = 5
$$\frac{p}{4}$$ x 4 = 5 x 4
∴ p = 20

(d) $$\frac{-p}{3}$$ x 3 = 5 x 3
-p = 15
∴ p = -15

(e) $$\frac{3 p}{4}$$ = 6
$$\frac{3 p}{4}$$ x 4 = 6 x 4
3p = 24
$$\frac{3 p}{3}=\frac{24}{3}$$
∴ p = 8

(f) 3s = -9
$$\frac{3 s}{3}=\frac{-9}{3}$$
∴ s = -3

(g) 3s+ 12 = 0
3s = -12
$$\frac{3 s}{3}=\frac{-12}{3}$$
∴ s = -4

(h) 3s = 0
or $$\frac{3 s}{3}=\frac{0}{3}$$
∴ s = 0

(i) 2q = 6
or $$\frac{2 q}{2}=\frac{6}{2}$$
∴ q = 3

(j) 2q – 6 = 6
or, 2q = +6
$$\frac{2 q}{2}=\frac{6}{2}$$
∴ q = 3

(k) 2q + 6 = 0
or 2q = -6
$$\frac{2 q}{2}=\frac{-6}{2}$$
∴ q = -3

(l) 2q + 6 = 12
or 2q = 12 – 6
$$\frac{2 q}{2}=\frac{6}{2}$$
∴ q = 3.