Question

graph the inequality.
$y \leq 3x + 2$

Explanation:

Step1: Graph the boundary line

The inequality is \( y \leq 3x + 2 \). First, graph the boundary line \( y = 3x + 2 \). This is a linear equation in slope - intercept form (\( y=mx + b \), where \( m = 3 \) (slope) and \( b = 2 \) (y - intercept)). Plot the y - intercept at \( (0,2) \). Then, use the slope \( m=\frac{3}{1} \) to find another point. From \( (0,2) \), move up 3 units and right 1 unit to get the point \( (1,5) \). Since the inequality is \( \leq \), the boundary line should be a solid line (because the inequality includes equality).

Step2: Shade the region

To determine which side of the line to shade, we can test a point not on the line. A common test point is the origin \( (0,0) \). Substitute \( x = 0 \) and \( y = 0 \) into the inequality \( y\leq3x + 2 \):
\( 0\leq3(0)+2 \)
\( 0\leq2 \), which is a true statement. So, we shade the region that includes the origin (the region below the line \( y = 3x+2 \) since the origin is below the line and satisfies the inequality).

Answer:

1. Draw a solid line for \( y = 3x + 2 \) (passing through \( (0,2) \) and \( (1,5) \) etc.).
2. Shade the region below (and including) the line \( y = 3x + 2 \).