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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.2y + 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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