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