Solve the following system of simultaneous linear equations by the substitution method:

5x + 4y - 4 = 0

x - 20 = 12y

Given,

5x + 4y - 4 = 0 ……(i)

x - 20 = 12y ……..(ii)

From eqn. (ii) we get,

x = 12y + 20 …….(iii)

Substituting value of x from eqn. (iii) in eqn. (i) we get,

⟹ 5(12y + 20) + 4y - 4 = 0

⟹ 60y + 100 + 4y - 4 = 0

⟹ 64y + 96 = 0

⟹ 64y = -96

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

Substituting value of y in eqn. (iii) we get,

$x = 12 \times \dfrac{-3}{2} + 20 \\[1em] = 6 \times -3 + 20 \\[1em] = -18 + 20 \\[1em] = 2.$

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