Solve 5x+4y=4 and x-12y=20 by Elimination and Substitution Method
Elimination Method
5x+4y=4 —-(i)
x-12y=20 —(ii)
Multiplying Equation (i) by 3 we get,
15x +12y = 12 —(iii)
Now Adding Equations (ii) and (iii) we get,
x+15x=20+12
⇒ 16x = 32
⇒ x = $\frac{32}{16}$
⇒ x = 2
From Equation (i) we get,
5(2) +4y=4
⇒ 10 +4y =4
⇒ 4y=4-10
⇒ 4y=-6
⇒ y=$\frac{ – 6}{4}$
⇒ y = – \frac{3}{2}
Therefore Required Solutions are x = 2 , y = – \frac{3}{2}
Substitution Method
5x+4y=4 —-(i)
x-12y=20 —(ii)
From Equation (ii) we get,
x = 20+12y
Substituting the value of x in equation (i) we get,
5(20+12y)+4y=4
⇒ 100 +60y +4y=4
⇒ 100+64y =4
⇒ 64y = 4-100
⇒ 64y = -96
⇒ y =$\frac{ – 96}{64}$
⇒ y= \frac{ – 3}{2}
Now from equation (iii) we get,
$x = 20 + 12y$
$\Rightarrow x = 20 + 12\left( – \frac{3}{2}\right)$
$\Rightarrow x = 20 – 18$
$\Rightarrow x = 2$
Therefore Required Solutions are x = 2 , y = – \frac{3}{2}