# Linear Equations in One Variable : Exercise 2.1 (Mathematics NCERT Class 8th)

Solve the following equations.

Q.1 $x - 2 = 7$
Sol. Given, $x - 2 = 7$
Adding 2 on both the sides, we get,
$x - 2 + 2 = 7 + 2$
$x = 7 + 2$
$x = 9$

Q.2$y + 3 = 10$
Sol. Given, $y + 3 = 10$
Subtracting 3 on both the sides, we get,
$y + 3 - 3 = 10 - 3$
$y = 10 - 3$
$y = 7$

Q.3 $6 = z + 2$
Sol. Given, $6 = z + 2$
On rearranging the terms, we get,
$z + 2 = 6$
Subtracting 2 on both the sides, we get,
$z + 2 - 2 = 6 - 2$
$z = 4$

Q.4 ${3 \over 7} + x = {{17} \over 7}$
Sol. Given, ${3 \over 7} + x = {{17} \over 7}$
Subtracting ${3 \over 7}$on both the sides, we get,
${3 \over 7} + x - {3 \over 7} = {{17} \over 7} - {3 \over 7}$
$x = {{17 - 3} \over 7}$
$x = {{14} \over 7}$
$x = 2$

Q.5 $6x = 12$
Sol. Given, $6x = 12$
Dividing by 6 on both the sides, we get,
${{6x} \over 6} = {{12} \over 6}$
$x = {{12} \over 6}$
$x = 2$

Q.6 ${t \over 5} = 10$
Sol. Given, ${t \over 5} = 10$
Multiplying by 5 on both the sides, we get,
${t \over 5} \times 5 = 10 \times 5$
$t = 10 \times 5$
$t = 50$

Q.7 ${{2x} \over 3} = 18$
Sol. Given, ${{2x} \over 3} = 18$
Multiplying by 3 on both the sides, we get,
${{2x} \over 3} \times 3 = 18 \times 3$
$2x = 54$
Dividing by 2 on both the sides, we get,
${{2x} \over 2} = {{54} \over 2}$
$x = 27$

Q.8 $1.6 = {y \over {1.5}}$
Sol. Given, $1.6 = {y \over {1.5}}$
Multiplying 1.5 on both the sides, we get,
$1.6 \times 1.5 = {y \over {1.5}} \times 1.5$
$1.6 \times 1.5 = y$
$2.40 = y$
$y = 2.4$

Q.9 $7x - 9 = 16$
Sol. Given, $7x - 9 = 16$
Adding 9 on both the sides, we get,
$7x - 9 + 9 = 16 + 9$
$7x = 25$
Dividing by 7 on both the sides, we get,
${{7x} \over 7} = {{25} \over 7}$
$x = {{25} \over 7}$

Q.10 $14y - 8 = 13$
Sol. Given, $14y - 8 = 13$
Adding 8 on both the sides, we get,
$14y - 8 + 8 = 13 + 8$
$14y = 21$
Dividing by 14 on both the sides, we get,
${{14y} \over {14}} = {{21} \over {14}}$
$y = {3 \over 2}$

Q.11 17 + 6p = 9
Sol. Given, 17 + 6p = 9
Subtracting 17 on both the sides, we get,
17 + 6p – 17 = 9 – 17
6p = – 8
Dividing by 6 on both the sides, we get,
${{6p} \over 6} = - {8 \over 6}$
$p = - {4 \over 3}$

Q.12 ${x \over 3} + 1 = {7 \over {15}}$
Sol. Given, ${x \over 3} + 1 = {7 \over {15}}$
Transposing 1 from LHS to RHS, we get,
${x \over 3} = {7 \over {15}} - 1$
${x \over 3} = {{7 - 15} \over {15}}$
${x \over 3} = {{ - 8} \over {15}}$
Multiplying by 3 on both the sides, we get,
${x \over 3} \times 3 = {{ - 8} \over {15}} \times 3$
$x = {{ - 8} \over 5}$

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