Solve the given inequality graphically in a two-dimensional plane: $3x + 4y \leq 12$.

(N/A) To solve the inequality $3x + 4y \leq 12$ graphically,we first consider the corresponding equation $3x + 4y = 12$.
Find the intercepts of the line $3x + 4y = 12$:
If $x = 0$,then $4y = 12 \implies y = 3$. So,the point is $(0, 3)$.
If $y = 0$,then $3x = 12 \implies x = 4$. So,the point is $(4, 0)$.
Draw the line passing through $(0, 3)$ and $(4, 0)$. Since the inequality is $\leq$,the line is solid.
To determine the solution region,test the origin $(0, 0)$ in the inequality $3x + 4y \leq 12$:
$3(0) + 4(0) = 0 \leq 12$,which is true.
Therefore,the solution region is the half-plane containing the origin $(0, 0)$,which is the shaded region below the line $3x + 4y = 12$.