Solve the following simultaneous equations:

$\dfrac{7 - 4x}{3} = y$, 2x + 3y + 1 = 0

Given,

Equations : $\dfrac{7 - 4x}{3} = y$, 2x + 3y + 1 = 0

⇒ $y = \dfrac{7 - 4x}{3}$     ….(1)

Substituting value of y from equation (1) in 2x + 3y + 1 = 0, we get :

$\Rightarrow 2x + 3\Big(\dfrac{7 - 4x}{3}\Big) + 1 = 0$

⇒ 2x + (7 - 4x) + 1 = 0

⇒ 8 - 2x = 0

⇒ 2x = 8

⇒ x = $\dfrac{8}{2}$

⇒ x = 4.

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

$\Rightarrow y = \dfrac{7 - 4x}{3} \\[1em] \Rightarrow y = \dfrac{7 - 4(4)}{3} \\[1em] \Rightarrow y = \dfrac{7 - 16}{3} \\[1em] \Rightarrow y = -\dfrac{9}{3} \\[1em] \Rightarrow y = -3.$ Hence, x = 4, y = -3.