# Solve the following pair of equations by the elimination method and the substitution method :

Question.

Solve the following pair of equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

(iv) $\frac{\mathbf{x}}{\mathbf{z}}+\frac{\mathbf{2} \mathbf{y}}{\mathbf{3}}=-\mathbf{1}$ and $x-\frac{\mathbf{y}}{\mathbf{3}}=3$

Solution:

(i) Solution By Elimination Method:

x + y = 5 ...(i)

2x – 3y = 4 ...(ii)

Multiplying (i) by 3 and (ii) by 1 and adding we get 3(x + y) + 1 (2x – 3y) = 3 × 5 + 1 × 4

$\Rightarrow 3 x+3 y+2 x-3 y=19$

$\Rightarrow 5 x=19 \Rightarrow x=\frac{19}{5}$

From (i), substituting $x=\frac{19}{5}$, we get

$\frac{\mathbf{1 9}}{\mathbf{5}}+\mathrm{y}=5 \Rightarrow \mathrm{y}=5-\frac{\mathbf{1 9}}{\mathbf{5}} \Rightarrow \mathrm{y}=\frac{\mathbf{6}}{\mathbf{5}}$

Hence, $x=\frac{19}{5}, y=\frac{6}{5}$

(i) Solution By Substitution Method :

x + y = 5 ...(i)

2x – 3y = 4 ...(ii)

From (i), y = 5 – x ...(iii)

Substituting y from (iii) in (ii), 2x – 3(5 – x) =4

$\Rightarrow 2 x-15+3 x=4$

$\Rightarrow 5 x=19 \Rightarrow x=\frac{19}{5}$

Then from (iii), $y=5-\frac{19}{5} \Rightarrow y=\frac{6}{5}$

Hence, $x=\frac{19}{5}, y=\frac{6}{5}$

(ii) Solution by elimination method

3x + 4y = 10 ...(i)

2x – 2y = 2 ...(ii)

multiplying (ii) equation by 2, we get

4x – 4y = 4 ...(iii)

Add equation (i) and (iii), we get

7x = 14

$\Rightarrow x=2$

Substituting, x = 2 in (i), we get

3 × 2 + 4 × y = 10

$\Rightarrow 4 y=4$

$\Rightarrow y=1$

Hence, x = 2, y = 1

(ii) Solution by substitution method

3x + 4y = 10 ...(i)

2x – 2y = 2 ...(ii)

From (ii), $y=\frac{2 x-2}{2}=x-1$...(iii)

Substituting, y = x – 1 in (i), we get

3x + 4 (x – 1) = 10

$\Rightarrow 3 x+4 x-4=10$

$\Rightarrow 7 x=14$

x = 2

Then from (iii)

y = 2 – 1 = 1

Hence, x = 2, y = 1

(iii) Solution by elimation method

3x – 5y = 4 ...(i)

9x = 2y + 7 ...(ii)

Multiplying (i) equation by 3, we get

9x – 15y = 12 ...(iii)

Subtracting (iii) from (ii), we get

9x – 9x + 15y = 2y + 7 – 12

$\Rightarrow 15 y-2 y=7-12$

13y = – 5

$y=\frac{-5}{13}$

From (i) substituting value of $y=\frac{-\mathbf{5}}{\mathbf{1 3}}$

$3 x=5 \times\left(\frac{-5}{13}\right)+4$

$\Rightarrow 3 x=\frac{-25}{13}+7$

$\Rightarrow 3 x=\frac{-25+52}{13}$

$3 x=\frac{27}{13}$

$x=\frac{9}{13}$

Hence, $y=\frac{-5}{13}, x=\frac{9}{13}$

(iii) Solution by substitution method

3x – 5y = 4 ...(i)

9x = 2y + 7 ...(ii)

From (i)

$\mathrm{x}=\frac{4+5 \mathrm{y}}{3}$ ...(iii)

Substuting $x=\frac{\mathbf{4}+\mathbf{5} \mathbf{y}}{\mathbf{3}}$ in (ii)

$9 \times \frac{4+5 y}{3}=2 y+7$

12 + 15y = 2y + 7

$y=\frac{-5}{13}$

from (iii)

$x=\frac{4+5\left(\frac{-5}{13}\right)}{3}=\frac{27}{39}$

Hence, $y=\frac{-5}{13}, x=\frac{9}{13}$

(iv) Solution by elimination method

$\frac{x}{2}+\frac{2 y}{3}=-1$ ...(i)

$x-\frac{y}{3}=3$ ...(ii)

Multiplying (ii), we get

$2 x-\frac{2 y}{3}=6$ ...(iii)

Adding (i) and (iii), we get

$2 x+\frac{x}{z}=-1+6$

$\Rightarrow \frac{5 x}{2}=5$

$\Rightarrow x=2$

From (ii) substituting x = 2, in equation (ii), we get

$\Rightarrow 2-\frac{\mathbf{y}}{\mathbf{3}}=3$

$\Rightarrow-1=\frac{\mathbf{y}}{\mathbf{3}}$

$\Rightarrow y=-3$

Hence, x = 2, y = – 3

(iv) Solution by substitution method

$\frac{x}{2}+\frac{2 y}{3}=-1$ ...(i)

$x-\frac{y}{3}=3$ ...(ii)

from (ii), $x=3+\frac{y}{3}$ ...(iii)

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

$\frac{3+\frac{y}{3}}{2}+\frac{2 y}{3}=-1$

$\Rightarrow \frac{3}{2}+\frac{y}{6}+\frac{2 y}{3}=-1$

$\Rightarrow \frac{\mathbf{y}+\mathbf{4} \mathbf{y}}{\mathbf{6}}=-1-\frac{\mathbf{3}}{\mathbf{2}}$

$\Rightarrow \frac{5 \mathbf{y}}{6}=\frac{-5}{2}$

$\Rightarrow y=-3$

Substituting y = – 3 in equation (ii), we get

$\Rightarrow x-\frac{(-3)}{3}=3$

$\Rightarrow x+1=3$

$\Rightarrow x=2$

Hence, x = 2, y = – 3