Question

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

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

\\)

Explanation:

Step1: Analyze \( y < 1 \)

The inequality \( y < 1 \) represents the region below the horizontal dashed line \( y = 1 \) (dashed because the inequality is strict, \( < \), so the line itself is not included).

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

The inequality \( y \leq 2x - 1 \) represents the region below or on the line \( y = 2x - 1 \). The line \( y = 2x - 1 \) has a slope of \( 2 \) and a \( y \)-intercept of \( -1 \). Since the inequality is \( \leq \), the line is solid (included in the region).

Step3: Find the intersection region

To find the solution to the system, we need the region that satisfies both inequalities. So we look for the area that is below \( y = 1 \) (dashed line) and below or on \( y = 2x - 1 \) (solid line).

Now, let's analyze the graphs:

• For \( y < 1 \), the region is below the dashed line \( y = 1 \).
• For \( y \leq 2x - 1 \), the region is below or on the line \( y = 2x - 1 \).

Looking at the three graphs, the second graph (middle one) has:

• The region below the dashed line \( y = 1 \) (blue area below \( y = 1 \)) and
• The region below or on \( y = 2x - 1 \) (purple area below or on the line \( y = 2x - 1 \)), and the intersection of these two regions is the area that is both below \( y = 1 \) and below or on \( y = 2x - 1 \), which matches the solution to the system.

Answer:

The middle graph (the second one among the three given graphs) is the graph of the system of linear inequalities \(

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

\).