Question

find the graph of this system of linear inequalities. \\(\

$$\begin{cases} y \\leq -2x + 4 \\\\ y \\leq 3x - 6 \\end{cases}$$

\\)

Explanation:

Step1: Analyze \( y \leq -2x + 4 \)

The line \( y = -2x + 4 \) has a slope of \(-2\) and y - intercept \( 4 \). Since the inequality is \( y \leq -2x + 4 \), we draw a solid line (because of the "equal to" part in \( \leq \)) and shade below the line.

Step2: Analyze \( y \leq 3x - 6 \)

The line \( y = 3x - 6 \) has a slope of \( 3 \) and y - intercept \(-6\). Since the inequality is \( y \leq 3x - 6 \), we draw a solid line and shade below the line.

Step3: Find the intersection of the two regions

To find the intersection of the regions defined by \( y \leq -2x + 4 \) and \( y \leq 3x - 6 \), we first find the point of intersection of the two lines \( y=-2x + 4\) and \( y = 3x-6\).
Set \(-2x + 4=3x - 6\)
\(-2x-3x=-6 - 4\)
\(-5x=-10\)
\(x = 2\)
Substitute \(x = 2\) into \(y=-2x + 4\), we get \(y=-2\times2 + 4=0\). So the two lines intersect at \((2,0)\).
The region that satisfies both inequalities is the region that is below both lines. We check the three graphs:

• For the first graph: The regions do not seem to be the intersection of the two lower - shaded regions.
• For the second graph: The purple region is the intersection of the region below \(y=-2x + 4\) (blue - shaded region) and the region below \(y = 3x-6\) (orange - shaded region) at the intersection point \((2,0)\) and below.
• For the third graph: The regions are not the correct intersection of the two lower - shaded regions.

Answer:

The second graph (the one with blue, orange, and purple regions where the purple region is the intersection of the regions below \(y=-2x + 4\) and \(y = 3x-6\))