What is the area (in unit squares) of the region enclosed by the graphs of the equations 2x - 3y + 6 = 0, 4x + y = 16 and y = 0?

Answer: Option A

2x - 3y + 6 = 0
y = 0
⇒ 2x - 3 × 0 = -6
2x = -6
x = -3
y = 0
y = 0 ⇒ 4x + 0 = 16
x = 4 ; y = 0
$$\begin{array}{rl}& 2x-3y=-6\ast 2\\ & \underline{4x+y=16}\ast 1\end{array}$$
4x - 6y = -12 . . . . . . (i)
4x + y = 16 . . . . . . (ii)
Solve equation (i) and (ii)
y = 4 ; x = 3

$$\Delta \text{ABC}=\frac{1}{2}\times \left(4+3\right)\times 4=14$$