Question

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

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

\\)

Explanation:

Step1: Analyze \( y < 2x - 2 \)

The inequality \( y < 2x - 2 \) is a linear inequality. The boundary line is \( y = 2x - 2 \), which has a slope of \( 2 \) and a \( y \)-intercept of \( -2 \). Since the inequality is \( y < 2x - 2 \), the line should be dashed (because the inequality is strict, \( < \) not \( \leq \)) and we shade the region below the line.

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

The inequality \( y \geq x - 1 \) has a boundary line \( y = x - 1 \), with a slope of \( 1 \) and a \( y \)-intercept of \( -1 \). Since the inequality is \( \geq \), the line should be solid (because equality is included) and we shade the region above the line.

Step3: Find the Intersection Region

We need to find the region that satisfies both inequalities. So we look for the area that is below the dashed line \( y = 2x - 2 \) and above the solid line \( y = x - 1 \).

Now let's analyze the graphs:

• For \( y < 2x - 2 \) (dashed line, shade below), and \( y \geq x - 1 \) (solid line, shade above).
• Looking at the three graphs, the third graph (the one with blue, purple, and orange regions) has the dashed line \( y = 2x - 2 \) (blue dashed line) and the solid line \( y = x - 1 \) (red solid line). The region that is below the dashed line and above the solid line is the purple region (or the region that is the intersection of the two shaded areas). Wait, actually, let's re - check:

Wait, maybe I mis - labeled. Let's take a point in the intersection region. Let's pick \( x = 2 \). For \( y < 2x - 2 \), when \( x = 2 \), \( y < 2(2)-2=2 \). For \( y \geq x - 1 \), when \( x = 2 \), \( y \geq 2 - 1 = 1 \). So a point like \( (2,1.5) \) should be in the solution region.

Looking at the three graphs:

• First graph: The dashed line is \( y = 2x - 2 \), solid line \( y=x - 1 \). The orange region is below \( y = x - 1 \) (since at \( x = 0 \), \( y=x - 1=-1 \), orange is below that), blue is below \( y = 2x - 2 \) (at \( x = 0 \), \( y = 2x - 2=-2 \), blue is below that). Not the intersection.
• Second graph: The dashed line is \( y = 2x - 2 \), solid line \( y=x - 1 \). The purple region is above \( y = 2x - 2 \) (at \( x = 0 \), \( y = 2x - 2=-2 \), purple is above that), which does not satisfy \( y < 2x - 2 \).
• Third graph: The dashed line is \( y = 2x - 2 \), solid line \( y=x - 1 \). The orange region is above \( y=x - 1 \) (at \( x = 0 \), \( y=x - 1=-1 \), orange is above that) and below \( y = 2x - 2 \) (at \( x = 0 \), \( y = 2x - 2=-2 \), orange is above \( - 2 \) but wait, no. Wait, when \( x = 3 \), \( y = 2x - 2 = 4 \), \( y=x - 1=2 \). The orange region at \( x = 3 \) is between \( y = 2 \) and \( y = 4 \)? No, wait the third graph: the solid line is \( y=x - 1 \) (red), dashed line \( y = 2x - 2 \) (blue). The orange region is above \( y=x - 1 \) (since for \( x = 2 \), \( y=x - 1 = 1 \), orange is above \( y = 1 \)) and below \( y = 2x - 2 \) (for \( x = 2 \), \( y = 2x - 2=2 \), orange is below \( y = 2 \)). So the point \( (2,1.5) \) is in the orange region of the third graph. Also, the boundary lines: \( y = 2x - 2 \) is dashed (correct for \( < \)) and \( y=x - 1 \) is solid (correct for \( \geq \)).

So the correct graph is the third one (the right - most graph among the three).

Answer:

The correct graph is the third (right - most) graph, which shows the region above the solid line \( y = x - 1 \) and below the dashed line \( y = 2x - 2 \).