Question

graph the inequality.
-3x + 4y > 16

Explanation:

Step1: Rewrite the inequality in slope - intercept form

We start with the inequality \(-3x + 4y>16\). We want to solve for \(y\) to get it in the form \(y = mx + b\) (slope - intercept form).
Add \(3x\) to both sides of the inequality: \(4y>3x + 16\).
Then divide each term by 4: \(y>\frac{3}{4}x + 4\).

Step2: Graph the boundary line

The boundary line for the inequality \(y>\frac{3}{4}x + 4\) is the line \(y=\frac{3}{4}x + 4\). Since the inequality is \(y>\frac{3}{4}x + 4\) (not \(y\geq\frac{3}{4}x + 4\)), the boundary line should be a dashed line.
To find two points on the line \(y=\frac{3}{4}x + 4\):

• When \(x = 0\), \(y=\frac{3}{4}(0)+4=4\). So one point is \((0,4)\).
• When \(x = 4\), \(y=\frac{3}{4}(4)+4=3 + 4=7\). So another point is \((4,7)\). Plot these two points and draw a dashed line through them.

Step3: Determine the region to shade

We 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 \(y>\frac{3}{4}x + 4\):
\(0>\frac{3}{4}(0)+4\), which simplifies to \(0 > 4\). This is false.
So we shade the region that does not contain the origin. In other words, we shade the region above the dashed line \(y=\frac{3}{4}x + 4\).

(Note: Since this is a graphing problem, the final answer is the graph with the dashed line \(y = \frac{3}{4}x+4\) and the region above the line shaded. If we were to describe the steps for graphing, the above steps are the key steps. If we consider the process of getting the inequality into slope - intercept form and determining the boundary and shading, the main steps are as above.)

Answer:

The graph of the inequality \(-3x + 4y>16\) (or \(y>\frac{3}{4}x + 4\)) has a dashed line with slope \(\frac{3}{4}\) and \(y\) - intercept \(4\), and the region above the dashed line is shaded.