Solve the following simultaneous equations:

5x + 4y = 4, x - 12y = 20

Given,

Equations : 5x + 4y = 4, x - 12y = 20

⇒ x - 12y = 20

⇒ x = 20 + 12y     ….(1)

Substituting value of x from equation (1) in 5x + 4y = 4, we get :

⇒ 5(20 + 12y) + 4y = 4

⇒ 100 + 60y + 4y = 4

⇒ 100 + 64y = 4

⇒ 64y = 4 - 100

⇒ 64y = -96

⇒ y = $-\dfrac{96}{64} = -\dfrac{3}{2}$.

Substituting value of y in equation (1), we get :

⇒ x = 20 + $12 \times -\Big(\dfrac{3}{2}\Big)$

⇒ x = 20 + 6(-3)

⇒ x = 20 - 18

⇒ x = 2.

Hence, x = 2, y = $-\dfrac{3}{2}$.