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

Answer:

(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

Answer:

(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)

Answer:

(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

Answer:

(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 = \(\frac{19}{7}\)