Answer
Answer $D$
Work Step by Step
Using the graph we will determine the equations of the two lines.
The "red" line has the intercepts $(3,0)$ and $(0,3)$. Its slope is:
$$m_1=\dfrac{3-0}{0-3}=-1.$$
Determine the equation of the "red" line using the slope $m_1$ and one of the intercepts, let's say $(0,3)$:
$$\begin{align*}
y-3&=-(x-0)\\
y&=-x+3\\
x+y&=3.
\end{align*}$$
The "blue" line has the intercepts $(4,0)$ and $(0,-4)$. Its slope is:
$$m_2=\dfrac{-4-0}{0-4}=1.$$
Determine the equation of the "blue" line using the slope $m_2$ and one of the intercepts, let's say $(0,-4)$:
$$\begin{align*}
y-(-4)&=x-0\\
-x+y&=-4.
\end{align*}$$
The solution of the inequality involving the "red" line is the half plane below the line, so the inequality is:
$$x+y<3.\tag1$$
The solution of the inequality involving the "blue" line is the half plane above the line, so the inequality is:
$$-x+y>-4.\tag2$$
Using inequalities $(1)$ and $(2)$ we find the system of inequalities:
$$\begin{cases}
x+y<3\\
-x+y>-4.
\end{cases}$$
The correct answer is Answer $D$.
