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}

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}