Solve the following simultaneous equations:

4x - 3y = 8, 18x - 3y = 29

Given,

Equations : 4x - 3y = 8, 18x - 3y = 29

⇒ 4x - 3y = 8

⇒ 4x = 3y + 8

⇒ x = $\dfrac{3y + 8}{4}$     ….(1)

Substituting value of x from equation (1) in 18x - 3y = 29, we get :

$\Rightarrow 18\Big(\dfrac{3y + 8}{4}\Big) - 3y = 29 \\[1em] \Rightarrow 9\Big(\dfrac{3y + 8}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72}{2}\Big) - 3y = 29 \\[1em] \Rightarrow \Big(\dfrac{27y + 72 - 6y}{2}\Big) = 29 \\[1em] \Rightarrow 21y + 72 = 29 \times 2 \\[1em] \Rightarrow 21y + 72 = 58 \\[1em] \Rightarrow 21y = 58 - 72 \\[1em] \Rightarrow 21y = -14 \\[1em] \Rightarrow y = \dfrac{-14}{21} \\[1em] \Rightarrow y = -\dfrac{2}{3}.$

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

$\Rightarrow x = \dfrac{3y + 8}{4} \\[1em] \Rightarrow x = \dfrac{3 \Big(\dfrac{-2}{3}\Big) + 8}{4} \\[1em] \Rightarrow x = \dfrac{-2 + 8}{4} \\[1em] \Rightarrow x = \dfrac{6}{4} \\[1em] \Rightarrow x = \dfrac{3}{2}.$

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