Simple Equations Ex-4.2

NCERT solutions for Maths Simple Equations 

NCERT Solutions for Class 7 Maths Simple Equations

Class –VII Mathematics (Ex. 4.2)

Question 1.Give first the step you will use to separate the variable and then solve the equations:

(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) $y + 4 = – 4$

Answer:

(a) $x – 1 = 0$ $\Rightarrow$ $x – 1 + 1 = 0 + 1$ [Adding 1 both sides]

$\Rightarrow$ $x = 1$

(b) $x + 1 = 0$ $\Rightarrow$ $x + 1 – 1 = 0 – 1$ [Subtracting 1 both sides]

$\Rightarrow$ $x = – 1$

(c) $x – 1 = 5$ $\Rightarrow$ $x – 1 + 1 = 5 + 1$ [Adding 1 both sides]

$\Rightarrow$ $x = 6$

(d) $x + 6 = 2$ $\Rightarrow$ $x + 6 – 6 = 2 – 6$ [Subtracting 6 both sides]

$\Rightarrow$ $x = – 4$

(e) $y – 4 = – 7$ $\Rightarrow$ $y – 4 + 4 = – 7 + 4$ [Adding 4 both sides]

$\Rightarrow$ $y = – 3$

(f) $y – 4 = 4$ $\Rightarrow$ $y – 4 + 4 = 4 + 4$ [Adding 4 both sides]

$\Rightarrow$ $y = 8$

(g) $y + 4 = 4$ $\Rightarrow$ $y + 4 – 4 = 4 – 4$ [Subtracting 4 both sides]

$\Rightarrow$ $y = 0$

(h) $y + 4 = – 4$ $\Rightarrow$ $y + 4 – 4 = – 4 – 4$ [Subtracting 4 both sides]

$\Rightarrow$ $y = – 8$

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NCERT Solutions for Class 7 Maths Exercise 4.2

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

(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) $20t = – 10$

Answer:

(a) $3l = 42$ $\Rightarrow$ $\frac{{3l}}{3} = \frac{{42}}{3}$ [Dividing both sides by 3]

$\Rightarrow$ $l = 14$

(b) $\frac{b}{2} = 6$ $\Rightarrow$ $\frac{b}{2} \times 2 = 6 \times 2$ [Multiplying both sides by 2]

$\Rightarrow$ $b = 12$

(c) $\frac{p}{7} = 4$ $\Rightarrow$ $\frac{p}{7} \times 7 = 4 \times 7$ [Multiplying both sides by 7]

$\Rightarrow$ $p = 28$

(d) $4x = 25$ $\Rightarrow$ $\frac{{4x}}{4} = \frac{{25}}{4}$ [Dividing both sides by 4]

$\Rightarrow$ $x = \frac{{25}}{4}$

(e) $8y = 36$ $\Rightarrow$ $\frac{{8y}}{8} = \frac{{36}}{8}$ [Dividing both sides by 8]

$\Rightarrow$ $y = \frac{9}{2}$

(f) $\frac{z}{3} = \frac{5}{4}$ $\Rightarrow$ $\frac{z}{3} \times 3 = \frac{5}{4} \times 3$ [Multiplying both sides by 3]

$\Rightarrow$ $z = \frac{{15}}{4}$

(g) $\frac{a}{5} = \frac{7}{{15}}$ $\Rightarrow$ $\frac{a}{5} \times 5 = \frac{7}{{15}} \times 5$ [Multiplying both sides by 5]

$\Rightarrow$ $a = \frac{7}{3}$

(h) $20t = – 10$ $\Rightarrow$ $\frac{{20t}}{{20}} = \frac{{ – 10}}{{20}}$ [Dividing both sides by 20]

$\Rightarrow$ $t = \frac{{ – 1}}{2}$

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NCERT Solutions for Class 7 Maths Exercise 4.2

Question 3.Give first the step you will use to separate the variable and then solve the equations

(a) $3n – 2 = 46$

(b) $5m + 7 = 17$

(c) $\frac{{20p}}{3} = 40$

(d) $\frac{{3p}}{{10}} = 6$

Answer:

(a) $3n – 2 = 46$

Step I: $3n – 2 + 2 = 46 + 2$ $\Rightarrow$ $3n = 48$

[Adding 2 both sides]

Step II: $\frac{{3n}}{3} = \frac{{48}}{3}$ $\Rightarrow$ $n = 16$ [Dividing both sides by 3]

(b) $5m + 7 = 17$

Step I: $5m + 7 – 7 = 17 – 7$ $\Rightarrow$ $5m = 10$ [Subtracting 7 both sides]

Step II: $\frac{{5m}}{5} = \frac{{10}}{5}$ $\Rightarrow$$m = 2$ [Dividing both sides by 5]

(c) $\frac{{20p}}{3} = 40$

Step I: $\frac{{20p}}{3} \times 3 = 40 \times 3$ $\Rightarrow$ $20p = 120$ [Multiplying both sides by 3]

Step II: $\frac{{20p}}{{20}} = \frac{{120}}{{20}}$ $\Rightarrow$ $p = 6$ [Dividing both sides by 20]

(d) $\frac{{3p}}{{10}} = 6$

Step I: $\frac{{3p}}{{10}} \times 10 = 6 \times 10$ $\Rightarrow$ $3p = 60$ [Multiplying both sides by 10]

Step II: $\frac{{3p}}{3} = \frac{{60}}{3}$ $\Rightarrow$ $p = 20$ [Dividing both sides by 3]

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NCERT Solutions for Class 7 Maths Exercise 4.2

Question 4.Solve the following equation:

(a) $10p = 100$

(b) $10p + 10 = 100$

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

(d) $\frac{{ – p}}{3} = 5$

(e) $\frac{{3p}}{4} = 6$

(f) $3s = – 9$

(g) $3s + 12 = 0$

(h) $3s = 0$

(i) $2q = 6$

(j) $2q – 6 = 0$

(k) $2q + 6 = 0$

(l) $2q + 6 = 12$

Answer:

(a) $10p = 100$ $\Rightarrow$ $\frac{{10p}}{{10}} = \frac{{100}}{{10}}$ [Dividing both sides by 10]

$\Rightarrow$ $p = 10$

(b) $10p + 10 = 100$ $\Rightarrow$ $10p + 10 – 10 = 100 – 10$ [Subtracting both sides 10]

$\Rightarrow$ $10p = 90$ $\Rightarrow$ $\frac{{10p}}{{10}} = \frac{{90}}{{10}}$ [Dividing both sides by 10]

$\Rightarrow$ $p = 9$

(c) $\frac{p}{4} = 5$ $\Rightarrow$ $\frac{p}{4} \times 4 = 5 \times 4$ [Multiplying both sides by 4]

$\Rightarrow$ $p = 20$

(d) $\frac{{ – p}}{3} = 5$ $\Rightarrow$ $\frac{{ – p}}{3} \times \left( { – 3} \right) = 5 \times \left( { – 3} \right)$ [Multiplying both sides by – 3]

$\Rightarrow$ $p = – 15$

(e) $\frac{{3p}}{4} = 6$ $\Rightarrow$ $\frac{{3p}}{4} \times 4 = 6 \times 4$ [Multiplying both sides by 4]

$\Rightarrow$ $3p = 24$ $\Rightarrow$ $\frac{{3p}}{3} = \frac{{24}}{3}$ [Dividing both sides by 3]

$\Rightarrow$ $p = 8$

(f) $3s = – 9$ $\Rightarrow$ $\frac{{3s}}{3} = \frac{{ – 9}}{3}$ [Dividing both sides by 3]

$\Rightarrow$ $s = – 3$

(g) $3s + 12 = 0$ $\Rightarrow$ $3s + 12 – 12 = 0 – 12$ [Subtracting both sides 10]

$\Rightarrow$ $3s = – 12$ $\Rightarrow$ $\frac{{3s}}{3} = \frac{{ – 12}}{3}$ [Dividing both sides by 3]

$\Rightarrow$ $s = – 4$

(h) $3s = 0$ $\Rightarrow$ $\frac{{3s}}{3} = \frac{0}{3}$ [Dividing both sides by 3]

$\Rightarrow$ $s = 0$

(i) $2q = 6$ $\Rightarrow$ $\frac{{2q}}{2} = \frac{6}{2}$ [Dividing both sides by 2]

$\Rightarrow$ $q = 3$

(j) $2q – 6 = 0$ $\Rightarrow$ $2q – 6 + 6 = 0 + 6$ [Adding both sides 6]

$\Rightarrow$ $2q = 6$ $\Rightarrow$ $\frac{{2q}}{2} = \frac{6}{2}$ [Dividing both sides by 2]

$\Rightarrow$ $q = 3$

(k) $2q + 6 = 0$ $\Rightarrow$ $2q + 6 – 6 = 0 – 6$ [Subtracting both sides 6]

$\Rightarrow$ $2q = – 6$ $\Rightarrow$ $\frac{{2q}}{2} = \frac{{ – 6}}{2}$ [Dividing both sides by 2]

$\Rightarrow$ $q = – 3$

(l) $2q + 6 = 12$ $\Rightarrow$ $2q + 6 – 6 = 12 – 6$ [Subtracting both sides 6]

$\Rightarrow$ $2q = 6$ $\Rightarrow$ $\frac{{2q}}{2} = \frac{6}{2}$ [Dividing both sides by 2]

$\Rightarrow$ $q = 3$

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