# Solve the following pair of linear equations by the substitution method.

Question.

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14, x – y = 4

(ii) $\mathrm{s}-\mathrm{t}=3, \frac{\mathbf{s}}{\mathbf{3}}+\frac{\mathbf{t}}{\mathbf{2}}=\mathbf{6}$

(iii) 3x – y = 3, 9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

(v) $\sqrt{\mathbf{2}} \mathbf{x}+\sqrt{\mathbf{3}} \mathbf{y}=\mathbf{0}, \sqrt{\mathbf{3}} \mathbf{x}-\sqrt{\mathbf{8}} \mathbf{y}=\mathbf{0}$

(vi) $\frac{3 x}{2}-\frac{5 y}{3}=-2, \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$

Solution:

(i) x + y = 14 ...(i)

x – y = 4 ...(ii)

From (ii) y = x – 4 ...(iii)

Substituting y from (iii) in (i), we get

x + x – 4 = 14

 2x = 18

 x = 9

Substituting x = 9 in (iii), we get

y = 9 – 4 = 5,

i.e, y = 5

x = 9, y = 5

(ii) s – t = 3 ...(i)

$\frac{s}{3}+\frac{t}{2}=6$...(ii)

From (i) s = t + 3 ...(iii)

Substituting s from (iii) in (ii), we get

$\frac{\mathbf{t}+\mathbf{3}}{\mathbf{3}}+\frac{\mathbf{t}}{\mathbf{2}}=\mathbf{6}$

 2(t + 3) + 3t = 36

 5t + 6 = 36

 t = 6

From (iii), s = 6 +3 = 9,

Hence, s = 9, t = 6

(iii) 3x – y = 3 .........(i)

9x – 3y = 9 .........(ii)

From (i) y = 3x – 3 ..........(iii)

Substituting y from (iii) in (ii), we get

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

It means, equation have infinite solutions.

(iv) 0.2x + 0.3y = 1.3 .......(i)

0.4x + 0.5y = 2.3 .......(ii)

From (i) $y=\frac{1.3-0.2 x}{0.3}$ ...(iii)

Substituting y from (iii) in (ii), we get

$0.4 x+0.5\left(\frac{1.3-0.2 x}{0.3}\right)=2.3$

$\Rightarrow 0.4 x+\frac{13}{6}-\frac{x}{3}=2.3$

$\Rightarrow \frac{2}{5} x-\frac{x}{3}=2.3-\frac{13}{6}$

$\Rightarrow \frac{x}{15}=\frac{4}{30}$

$\Rightarrow x=2$

Substituting x = 2 in (iii)

y = 3 × 2 – 3

Hence, y = 3

(v) $\sqrt{2} x+\sqrt{3} y=0$...(i)

$\sqrt{3} x-\sqrt{8} y=0$ ...(ii)

From (ii) $\mathrm{y}=\frac{\sqrt{\mathbf{3}} \mathbf{x}}{\sqrt{\mathbf{8}}}$ ...(iii)

Substituting y from (iii) in (i), we get

$\sqrt{2} x+\sqrt{3} \times \frac{\sqrt{3} x}{\sqrt{8}}=0$

$\Rightarrow \frac{4 \mathbf{x}+\mathbf{3 x}}{\sqrt{\mathbf{8}}}=0 \Rightarrow 7 \mathrm{x}=0$

$x=0$

Substituting x = 0 in (iii)

Hence, y = 0

(vi) $\frac{\mathbf{3 x}}{\mathbf{2}}-\frac{\mathbf{5 y}}{\mathbf{3}}=-2$ ....(i)

$\frac{x}{3}+\frac{y}{2}=\frac{13}{6}$...(ii)

From (i) $y=\frac{\frac{3 x}{2}+2}{5 / 3}=\frac{9 x+12}{10}$...(iii)

Substituting y from (iii) in (ii), we get

$\frac{x}{3}+\frac{9 x+12}{10 \times 2}=\frac{13}{6}$

$\Rightarrow \frac{x}{3}+\frac{9 x}{20}+\frac{3}{5}=\frac{13}{6}$

$\Rightarrow \frac{47 x}{60}=\frac{47}{30}$

x = 2

Substituting x = 2 in (iii)

$y=\frac{9 \times 2+12}{10}$

Hence, y = 3.