Question

3x + 4y < 12
-7x + 9y < 63
3x + 4y ≥ 12
-7x + 9y > 63
3x + 4y ≥ 12
-7x + 9y < 63
3x + 4y < 12
-7x + 9y ≥ 63

Explanation:

Step1: Analyze the inequality $3x + 4y<12$

The boundary - line of the inequality $3x + 4y = 12$ has a $y$ - intercept of $y = 3$ (when $x = 0$) and an $x$ - intercept of $x = 4$ (when $y = 0$). Since the inequality is $3x + 4y<12$, the region below the line $3x + 4y = 12$ is considered for this inequality.

Step2: Analyze the inequality $-7x+9y < 63$

The boundary - line of the inequality $-7x + 9y=63$ has a $y$ - intercept of $y = 7$ (when $x = 0$) and an $x$ - intercept of $x=-9$ (when $y = 0$). Since the inequality is $-7x + 9y<63$, the region below the line $-7x + 9y = 63$ is considered for this inequality.

Step3: Check the regions in the graph

By looking at the graph, we can see that the double - shaded region (the region that satisfies both inequalities simultaneously) is the region that is below the line $3x + 4y = 12$ and below the line $-7x + 9y = 63$. This corresponds to the system of inequalities

$$\begin{cases}3x + 4y<12\\-7x + 9y<63\end{cases}$$

Answer:

$$\begin{cases}3x + 4y<12\\-7x + 9y<63\end{cases}$$