Question

graph the following system of inequalities on the coordinate plane and shade the resulting solution region: \\(\

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

\\)

Explanation:

Step1: Graph \( y > 2x - 3 \)

First, graph the line \( y = 2x - 3 \). The slope \( m = 2 \) and the y - intercept \( b=-3 \). Since the inequality is \( y>2x - 3 \), the line should be dashed (because the inequality is strict, \( y\) is not equal to \( 2x - 3 \)). Then, to determine which side of the line to shade, we can use a test point. A common test point is \( (0,0) \). Substitute \( x = 0 \) and \( y = 0 \) into the inequality: \( 0>2(0)-3\), which simplifies to \( 0>-3 \). This is true, so we shade the region above the dashed line \( y = 2x - 3 \).

Step2: Graph \( y\leq -x + 4 \)

Next, graph the line \( y=-x + 4 \). The slope \( m=-1 \) and the y - intercept \( b = 4 \). Since the inequality is \( y\leq -x + 4 \), the line should be solid (because \( y\) can be equal to \( -x + 4 \)). Use the test point \( (0,0) \) again. Substitute \( x = 0 \) and \( y = 0 \) into the inequality: \( 0\leq-(0)+4\), which simplifies to \( 0\leq4 \). This is true, so we shade the region below (or on) the solid line \( y=-x + 4 \).

Step3: Find the solution region

The solution region for the system of inequalities is the region where the shading from both inequalities overlaps. So we look for the area that is above the dashed line \( y = 2x - 3 \) and below (or on) the solid line \( y=-x + 4 \).

Answer:

To graph the system:

1. For \( y>2x - 3 \), draw a dashed line with slope \( 2 \) and y - intercept \( -3 \), shade above it.
2. For \( y\leq -x + 4 \), draw a solid line with slope \( -1 \) and y - intercept \( 4 \), shade below (and on) it.
3. The overlapping shaded region (above \( y = 2x - 3 \) dashed line and below/on \( y=-x + 4 \) solid line) is the solution. (Graphing is best done on a coordinate plane with the two lines and the overlapping region shaded appropriately.)