Question

find the graph of this system of linear inequalities. \

$$\begin{cases} y \\leq -x + 1 \\\\ y \\geq 2x - 1 \\end{cases}$$

Explanation:

Step1: Analyze \( y \leq -x + 1 \)

The line \( y=-x + 1 \) has a slope of \(-1\) and a \( y \)-intercept at \( (0,1) \). Since the inequality is \( y\leq -x + 1 \), we draw a solid line (because the inequality is non - strict) and shade the region below the line.

Step2: Analyze \( y\geq 2x - 1 \)

The line \( y = 2x-1 \) has a slope of \( 2 \) and a \( y \)-intercept at \( (0,-1) \). Since the inequality is \( y\geq 2x - 1 \), we draw a solid line (because the inequality is non - strict) and shade the region above the line.

Step3: Find the intersection region

The solution to the system of inequalities is the region that is shaded for both inequalities. We can also test a point in the intersection region. Let's take the point of intersection of the two lines \( y=-x + 1 \) and \( y = 2x-1 \). Set \(-x + 1=2x-1\), then \( 3x=2\), \( x=\frac{2}{3}\), and \( y=-\frac{2}{3}+1=\frac{1}{3}\). Now, let's test the point \( (0,0) \) in both inequalities:

• For \( y\leq -x + 1 \): \( 0\leq0 + 1\), which is true.
• For \( y\geq 2x - 1 \): \( 0\geq0 - 1\), which is true. So \( (0,0) \) is in the solution region.

Looking at the three graphs, the third graph (the one with blue, orange, and purple regions where the purple region is the intersection) shows the region below \( y=-x + 1 \) and above \( y = 2x-1 \) (including the lines), and the point \( (0,0) \) is in the purple region.

Answer:

The third graph (the one with the purple region that is the intersection of the regions defined by \( y\leq -x + 1 \) and \( y\geq 2x - 1 \)) is the graph of the system of linear inequalities.