Question

find the graph of this system of linear inequalities. \

$$\begin{cases} x \\leq 3 \\\\ y > -x + 2 \\end{cases}$$

Explanation:

Step1: Analyze \( x \leq 3 \)

The inequality \( x \leq 3 \) represents all points to the left of and including the vertical line \( x = 3 \). So the region for this inequality is shaded to the left of \( x = 3 \) (the line \( x = 3 \) is solid because of the "less than or equal to" sign).

Step2: Analyze \( y > -x + 2 \)

The inequality \( y > -x + 2 \) is a linear inequality. The boundary line is \( y = -x + 2 \), which has a slope of \( -1 \) and a y - intercept of \( 2 \). Since the inequality is "greater than", the boundary line is dashed, and we shade the region above the line \( y=-x + 2 \).

Step3: Find the intersection

The solution to the system of inequalities is the region that is shaded both to the left of \( x = 3 \) (including \( x = 3 \)) and above the line \( y=-x + 2 \) (not including the line \( y=-x + 2 \)).

Looking at the three graphs:

• For the first graph: Check the shading for \( x\leq3 \) (left of \( x = 3 \)) and \( y > -x + 2 \) (above the line). The shading of the two inequalities should overlap in the region that satisfies both.
• For the second graph: The shading does not match the required regions for both inequalities.
• For the third graph: The shading also does not match the required regions for both inequalities.
• The first graph (the left - most one) has the correct shading: the region to the left of \( x = 3 \) (solid line at \( x = 3 \)) and above the dashed line \( y=-x + 2 \) (since the slope of \( y=-x + 2 \) is - 1, y - intercept 2, and the shading above the line and left of \( x = 3 \) is correct).

Answer:

The graph that represents the system \(

$$\begin{cases}x\leq3\\y > -x + 2\end{cases}$$

\) is the left - most graph (with the blue shading on the left of \( x = 3 \) and the orange shading above \( y=-x + 2 \) overlapping in the correct region).