Question

1. y > x + 1; y ≥ -x + 1

Explanation:

Step1: Analyze \( y > x + 1 \)

The inequality \( y > x + 1 \) is a linear inequality. The boundary line is \( y = x + 1 \) (slope \( 1 \), y - intercept \( 1 \)). Since the inequality is \( > \), the line is dashed, and we shade above the line.

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

The inequality \( y \geq -x + 1 \) has a boundary line \( y = -x + 1 \) (slope \( - 1 \), y - intercept \( 1 \)). Since the inequality is \( \geq \), the line is solid, and we shade above the line.

Step3: Graph the boundary lines

• For \( y = x + 1 \): Plot the y - intercept \( (0,1) \). Use the slope \( 1 \) (rise 1, run 1) to find another point, e.g., \( (1,2) \). Draw a dashed line through these points.
• For \( y=-x + 1 \): Plot the y - intercept \( (0,1) \). Use the slope \( - 1 \) (rise - 1, run 1) to find another point, e.g., \( (1,0) \). Draw a solid line through these points.

Step4: Shade the regions

• For \( y > x + 1 \), shade the region above the dashed line \( y = x + 1 \).
• For \( y \geq -x + 1 \), shade the region above the solid line \( y=-x + 1 \).
• The solution to the system of inequalities is the intersection of these two shaded regions. The intersection is the region that is above both lines, with the boundary of \( y=-x + 1 \) being included (solid line) and the boundary of \( y = x + 1 \) being excluded (dashed line). The two lines intersect at \( (0,1) \) (by solving \( x + 1=-x + 1\Rightarrow2x = 0\Rightarrow x = 0,y = 1\)).

Answer:

To graph the system \(

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

\):

1. Draw a dashed line for \( y = x + 1 \) (slope \( 1 \), y - intercept \( 1 \)) and shade above it.
2. Draw a solid line for \( y=-x + 1 \) (slope \( - 1 \), y - intercept \( 1 \)) and shade above it.
3. The solution region is the intersection of the two shaded regions (above both lines, including the solid - lined boundary \( y=-x + 1 \) and excluding the dashed - lined boundary \( y = x + 1 \)), with the lines intersecting at \( (0,1) \).