# NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.3

Question 1.
Solve the following equations:
(a) 2y + $$\frac{5}{2}=\frac{37}{2}$$
(b) 5t + 28 = 10
(c) $$\frac{\mathrm{a}}{5}$$ + 3 = 2
(d) $$\frac{\mathrm{q}}{4}$$ + 7 = 5
(e) $$\frac{5}{2}$$ x = -10
(f) $$\frac{5}{2}$$ x = $$\frac{25}{4}$$
(g) 7m + $$\frac{19}{2}$$ = 13
(h) 6z + 10 = -2
(i) $$\frac{31}{2}=\frac{2}{3}$$
(j) $$\frac{2 \mathrm{~b}}{3}$$ – 5 = 3
(a) we have
2y + $$\frac{5}{2}=\frac{37}{2}$$
Transposing $$\frac{5}{2}$$ from L.H.S to R.H.S
2y = $$\frac{37}{2}-\frac{5}{2}$$
2y = $$\frac{37-5}{2}$$
2y = $$\frac{32}{2}$$
2y = 16.

(b) we have
5t + 28 = 10
Transposing 28 from L.H.S to R.H.S
5t = 10 – 28
5t = – 18
Dividing both sides by 5, we get
$$\frac{5 \mathrm{t}}{5}=\frac{-18}{5}$$
t = $$\frac{-18}{5}$$
The solution is t = $$\frac{-18}{5}$$ or -3 $$\frac{3}{5}$$
Dividing both sides by 2, we get
$$\frac{2 y}{2}=\frac{16}{2}$$
y = 8
y = 8 is the required solution.

(c) we have a
$$\frac{\mathrm{a}}{5}$$ + 3 = 2
Transposing 3 from L.H.S to R.H.S
$$\frac{\mathrm{a}}{5}$$ = 2 – 3
$$\frac{\mathrm{a}}{5}$$ = -1
Multiplying both sides by 5, we get
$$\frac{\mathrm{a} \times 5}{5}$$ = -1 x 5
a = – 5
The solution is a = – 5.

(d) We have
$$\frac{\mathrm{q}}{4}$$ + 7 = 5
Transposing 7 to R.H.S
$$\frac{\mathrm{q}}{4}$$ = 5 – 7
$$\frac{\mathrm{q}}{4}$$ = -2
Multiplying both sides by 4, we get
$$\frac{\mathrm{q}}{4}$$ × 4 = – 2 × (4)
q = -8
The solution is q = – 8.

(e) We have
$$\frac{5}{2}$$ x = – 10
Multiplying both sides by 2, we get
$$\frac{5}{2}$$ x x 2 = -10 × 2
5x = – 20
Dividing both sides by 5, we get
$$\frac{5 x}{5}=\frac{-20}{5}$$
The solution is x = – 4.

(f) We have
$$\frac{5}{2}$$ x = $$\frac{25}{4}$$
Multiplying both sides by 2, we get
$$\frac{5}{2}$$ x × 2 = $$\frac{25}{4}$$ × 2
5x = $$\frac{25}{2}$$
Dividing both sides by 5, we get
$$\frac{5 x}{5}=\frac{25}{2 \times 5}$$
x = $$\frac{5}{2}$$
The solution is x= $$\frac{5}{2}$$ or x = 2 $$\frac{1}{2}$$.

(g) We have
7m + $$\frac{19}{2}$$ = 13
Transposing $$\frac{19}{2}$$ to R.H.S
7m = 13 – $$\frac{19}{2}$$
7m = $$\frac{26-19}{2}$$
7m = $$\frac{7}{2}$$

(h) We have
6z + 10 = – 2
Transposing 10 to R.H.S
6z = – 2 – 10
6z = – 12
Dividing both sides by 6, we get
$$\frac{6 z}{6}=\frac{-12}{6}$$
z = – 2
The solution is z = – 2.
Dividing both sides by 7, we get
$$\frac{7 \mathrm{~m}}{7}=\frac{7}{2 \times 7}$$
m = $$\frac{1}{2}$$
The solution is m = $$\frac{1}{2}$$ .

(i) We have
$$\frac{31}{2}=\frac{2}{3}$$
Multiplying both sides by 2, we get
$$\frac{31}{2}$$ x 2 = $$\frac{2}{3}$$ x 2
31 = $$\frac{4}{3}$$
Dividing both sides by 3, we get
$$\frac{31}{3}=\frac{4}{3 \times 3}$$
1 = $$\frac{4}{9}$$
The solution is 1 = $$\frac{4}{9}$$

(j) We have
$$\frac{2 \mathrm{~b}}{3}$$ – 5 = 3
Transposing – 5 to R.H.S
$$\frac{2 \mathrm{~b}}{3}$$ = 3 + 5
$$\frac{2 \mathrm{~b}}{3}$$ = 8
Multiplying both sides by 3, we get
$$\frac{2 \mathrm{~b}}{3}$$ × 3 = 8 × 3
2b = 24
Dividing both sides by 2, we get
$$\frac{2 b}{2}=\frac{24}{2}$$
b = 12
The solution is b = 12.

Question 2.
Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
(a) 2(x + 4) = 12
Dividing both sides by 2, we get
$$\frac{2(x+4)}{2}=\frac{12}{2}$$ ⇒ x + 4 = 6
Transposing 4 to R.H.S
x = 6 – 4 = 2

(b) 3(n – 5) = 21
Dividing both sides by 3, we get
$$\frac{3(n-5)}{3}=\frac{21}{3}$$
⇒ n – 5 = 7
Transposing – 5 to R.H.S
n = 7 + 5 = 12

(c) 3(n – 5) = -21
Dividing both sides by 3, we get
$$\frac{3(n-5)}{3}=\frac{-21}{3}$$
⇒ n – 5 = -7
Transposing – 5 to R.H.S
n = -7 + 5 = -2

(d) – 4(2 + x) =8
Dividing both sides by -4, we get
$$\frac{-4(2+x)}{-4}=\frac{8}{-4}$$
⇒ 2 + x = – 2
Transposing 2 to R.H.S
x = -2 -2 = -4

(e) 4(2 – x) = 8
Dividing both sides by 4, we get 4(2-x) _ 8
$$\frac{4(2-\mathrm{x})}{4}=\frac{8}{4}$$
⇒ 2 – x = 2
Transposing 2 to R.H.S
-x = 2 – 2 = 0
Multiplying both sides by (- 1), we get
-x × (-1) = 0 × (- 1)
x = 0

Question 3.
Solve the following equations:
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
(a) 4 = 5 (p – 2)
Interchanging the sides, we get 5 ( p – 2) =4
Dividing both sides by 5, we get
$$\frac{5(p-2)}{5}=\frac{4}{5}$$
⇒ p – 2 = $$\frac{4}{5}$$
Transposing -2 to R.H.S
p = $$\frac{4}{5}$$ + 2
= $$\frac{4+10}{5}=\frac{14}{5}$$ = 2$$\frac{4}{5}$$

(b) – 4 = 5(p – 2)
Interchanging the sides, we get }
5 ( p – 2) = -4
Dividing both sides by 5, we get
$$\frac{5(p-2)}{5}=\frac{-4}{5}$$
p – 2 = $$\frac{-4}{5}$$
Transposing – 2 to R.H.S
p = $$\frac{-4}{5}$$ + 2 = $$\frac{-4+10}{5}$$
= $$\frac{6}{5}$$ = 1$$\frac{1}{5}$$

(c) 16 = 4 + 3(t + 2)
Interchanging the sides, we get 4 + 3 (t + 2) = 16
Transposing 4 to R.H.S
3(t + 2) – 16 – 4 = 12
Dividing both sides by 3, we get
$$\frac{3(t+2)}{3}=\frac{12}{3}$$
⇒ t + 2 = 4
Transposing 2 to R.H.S
t = 4 – 2 = 2

(d) 4 + 5 (p – 1) = 34
Transposing 4 to R.H.S
5 (p – 1) = 34 – 4 = 3
Dividing both sides by 5, we get
$$\frac{5(\mathrm{p}-1)}{5}=\frac{30}{5}$$
p – 1 = 6
Transposing -1 to R.H.S
p = 6 + 1 = 7

(e) 0 = 16 + 4(m – 6)
Interchanging the sides we get
16 + 4(m – 6) = 0
Transposing 16 to R.H.S
4(m – 6) = 0 – 16 = – 16
Dividing both sides by 4 we get
$$\frac{4(\mathrm{~m}-6)}{4}=\frac{-16}{4}$$
⇒ m – 6 = – 4
Transposing – 6 to R.H.S
m = -4 + 6 = 2

Question 4.
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2
(a) Starting with x = 2
(1) Multiplying both sides by 3, we get
3x = 6
Subtracting 7 from both sides, we get
3x – 7 = 6 – 7
3x – 7 = – 1

(2) x = 2
Adding 5 on both sides, we get
x + 5 = 2 + 5
x + 5 = 7
Multiplying both sides by 4, we get
4(x + 5) = 7 × 4
4 (x + 5) = 28

(b) Starting with x = – 2
(1) x = – 2
Adding 11 to both sides, we get
x + 11 = – 2 + 11
x + 11 = 9

(2) x = – 2
Multiplying both sides by 5, we get
5x = – 10
Adding 3 on both sides, we get
5x + 3 = – 10 + 3
5x + 3 = – 7

(3) x = 2
Dividing both sides by 5, we get
$$\frac{x}{5}=\frac{2}{5}$$
Subtracting 2 from both sides, we get
$$\frac{\mathrm{x}}{5}$$ – 2 = $$\frac{2}{5}$$ -2
$$\frac{\mathrm{x}}{5}$$ – 2 = $$\frac{2-10}{5}$$
$$\frac{\mathrm{x}}{5}$$ – 2 = $$\frac{-8}{5}$$

(3) x = – 2
Dividing both sides by 7, we get
$$\frac{x}{7}=\frac{-2}{7}$$
Add 3 on both sides, we get
$$\frac{\mathrm{X}}{7}$$ + 3 = $$\frac{-2}{7}$$ + 3
= $$\frac{-2+21}{7}$$
$$\frac{\mathrm{X}}{7}$$ + 3 = $$\frac{19}{7}$$

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