Question

v. graph each of the inequalities: 13. $3x + 2y \leq 4$

Explanation:

Step1: Rewrite in slope - intercept form

We start with the inequality \(3x + 2y\leq4\). First, we solve for \(y\) to get it in the form \(y = mx + b\) (slope - intercept form).
Subtract \(3x\) from both sides: \(2y\leq - 3x + 4\).
Then divide each term by \(2\): \(y\leq-\frac{3}{2}x + 2\).

Step2: Graph the boundary line

The boundary line is \(y =-\frac{3}{2}x + 2\). This is a straight line with a slope of \(-\frac{3}{2}\) and a \(y\) - intercept of \(2\) (when \(x = 0\), \(y=2\)) and an \(x\) - intercept of \(\frac{4}{3}\) (when \(y = 0\), \(0=-\frac{3}{2}x + 2\), \(\frac{3}{2}x=2\), \(x=\frac{4}{3}\)). Since the inequality is \(\leq\), we draw a solid line (because the points on the line are included in the solution set).

Step3: Determine the shading 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 original inequality: \(3(0)+2(0)\leq4\), which simplifies to \(0\leq4\). This is a true statement. So we shade the region that includes the origin (the region below the line \(y =-\frac{3}{2}x + 2\) since the origin is below the line and satisfies the inequality).

Answer:

To graph \(3x + 2y\leq4\):

1. Rewrite as \(y\leq-\frac{3}{2}x + 2\).
2. Draw the solid line \(y =-\frac{3}{2}x + 2\) (slope \(-\frac{3}{2}\), \(y\) - intercept \(2\), \(x\) - intercept \(\frac{4}{3}\)).
3. Shade the region below (and including) the line (the region that contains the origin \((0,0)\) as it satisfies the inequality).