Solve: 2x-y/4=5 and x+3y/4 =-1 by Elimination and Substitution Method

Solve: 2x-y/4=5 and x+3y/4 =-1 by Elimination and Substitution Method.

Elimination Method

$2x – \frac{y}{4} = 5$ —-(i)

$x + \frac{3y}{4} = – 1$ —(ii)

Multiplying Equation (i) by 3 we get,

$6x – \frac{3y}{4} =15$ —(iii)

Adding equations (ii) and (iii) we get,

$x + \frac{3y}{4}+ 6x – \frac{3y}{4}$ = -1+15

⇒ 7x = 14

⇒ x = $\frac{14}{7}$

⇒ x = 2

Substituting the value of x in equation (i) we get,

$2x – \frac{y}{4} = 5$

$\Rightarrow 2(2) – \frac{y}{4} = 5$

$\Rightarrow 4 – \frac{y}{4} = 5$

$\Rightarrow 4 – 5 = \frac{y}{4}$

$\Rightarrow – 1 = \frac{y}{4}$

$\Rightarrow y = – 4$

Therefore Required Solutions are x = 2 and y= -4

Substitution Method

$2x – \frac{y}{4} = 5$ —-(i)

$x + \frac{3y}{4} = – 1$ —(ii)

From Equation (i) we get,

$2x – \frac{y}{4} = 5$

$\Rightarrow 2x – 5 = \frac{y}{4}$

$\Rightarrow 4(2x – 5) = y$

$\Rightarrow y = 8x – 20$ —-(iii)

Substituting the value of y in equation (ii) we get,

$x + \frac{3y}{4} = – 1$

$\Rightarrow x + \frac{3(8x – 20)}{4} = – 1$

$\Rightarrow \frac{4x + 24x – 60}{4} = – 1$

$\Rightarrow 28x – 60 = – 4$

$\Rightarrow 28x = 60 – 4$

$\Rightarrow 28x = 56$

$\Rightarrow x = \frac{56}{28}$

$\Rightarrow x = 2$

Now From Equation (iii) we get,

y=8(2)-20 [ as, x =2]

⇒ y = 16-20

⇒ y =-4

Therefore Required Solutions are x = 2 and y= -4

error: Content is protected !!