Class 7 Simple Equations Ex 4.2

 

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

1. Give first the step you will use to separate the variable and then solve the equation.

(a) x – 1 = 0

Solution:-

 x = 0 + 1

 x = 1

(b) x + 1 = 0

Solution:-

x  = 0 – 1

x = – 1

(c) x – 1 = 5

Solution:-

 x  = 5 + 1

 x = 6

(d) x + 6 = 2

Solution:-

 x  = 2 – 6

 x = – 4

(e) y – 4 = – 7

Solution:-

 y  = – 7 + 4

 y = – 3

(f) y – 4 = 4

Solution:-

 y = 4 + 4

 y = 8

(g) y + 4 = 4

Solution:-

 y  = 4 – 4

y = 0

(h) y + 4 = – 4

Solution:-

 y  = – 4 – 4

y = – 8

2. Give first the step you will use to separate the variable and then solve the equation.

(a) 3l = 42

Solution:-

\(l = {42\over3}\)

 l = 14

(b) \({b\over2} = 6\)

Solution:-

b= 6 × 2

b = 12

(c) \({p\over7} = 4\)

Solution:-

p= 4 × 7

p = 28

(d) 4x = 25

Solution:-

x = \(25\over4\)

(e) 8y = 36

Solution:-

\(y={36\over8}\)

\(y={9\over2}\)

(f) \(({z\over3}) = ({5\over4})\)

Solution:-

\(z = ({5\over4}) × 3\)

\( x = {15\over4}\)

(g) \(({a\over5}) = ({7\over15})\)

Solution:-

\(a = ({7\over15})× 5\)

a = \(7\over3\)

(h) 20t = – 10

Solution:-

\(t= {-10\over20}\)

\(x = {-1 \over 2}\)

3. Give the steps you will use to separate the variable and then solve the equation.

(a) 3n – 2 = 46

Solution:-

3n  = 46 + 2

3n = 48

\(n = {48\over3}\)

 n = 16

(b) 5m + 7 = 17

Solution:-

5m  = 17 – 7

5m = 10

\(m = {10\over5}\)

 m = 2

(c) \({20p\over3} = 40\)

Solution:-

20p= 40 × 3

20p = 120

\(p = {120\over20}\)

 p = 6

(d) 3p/10 = 6

Solution:-

3p = 6 × 10

3p = 60

\(p = {60\over3}\)

p = 20

4. Solve the following equations.

(a) 10p = 100

Solution:-

\(p = {100\over10}\)

= p = 10

(b) 10p + 10 = 100

Solution:-

10p  = 100 – 10

10p = 90

\(p = {90\over10}\)

p = 9

(c) \({p\over4} = 5\)

Solution:-

p= 5 × 4

p = 20

(d) \({– p\over3} = 5\)

Solution:-

\(– p = 5 × (- 3)\)

 p = – 15

(e) \({3p\over4} = 6\)

Solution:-

3p = 6 × 4

\(3p = 24\)

\(p = {24\over3}\)

p = 8

(f) 3s = – 9

Solution:-

\(s = {-9\over3}\)

s = -3

(g) 3s + 12 = 0

Solution:-

3s = 0 – 12

 3s = -12

\(s = {-12\over3}\)

s = – 4

(h) 3s = 0

Solution:-

\(s = {0\over3}\)

s = 0

(i) 2q = 6

Solution:-

\(q = {6\over 2}\)

q = 3

(j) 2q – 6 = 0

Solution:-

 2q  = 0 + 6

 2q = 6

\(2q = {6\over2}\)

q = 3

(k) 2q + 6 = 0

Solution:-

First, we have to subtract 6 from both sides of the equation.

2q + 6 – 6 = 0 – 6

2q = – 6

\(q = {– 6\over2}\)

q = – 3

(l) 2q + 6 = 12

Solution:-

2q= 12 – 6

2q = 6

\(q = {6\over 2}\)

q = 3