Question

graph the inequality on the axes below.
-2x + 3y ≥ 3

Explanation:

Step1: Rewrite in slope - intercept form

First, we start with the inequality \(-2x + 3y\geq3\). We want to solve for \(y\) to get it in the form \(y = mx + b\) (slope - intercept form).
Add \(2x\) to both sides of the inequality: \(3y\geq2x + 3\).
Then divide each term by 3: \(y\geq\frac{2}{3}x + 1\).

Step2: Graph the boundary line

The boundary line is \(y=\frac{2}{3}x + 1\). Since the inequality is \(\geq\), the boundary line should be a solid line (because the points on the line are included in the solution set).
To graph the line, we can find the \(y\) - intercept and the slope. The \(y\) - intercept \(b = 1\), so the line passes through the point \((0,1)\). The slope \(m=\frac{2}{3}\), which means from the point \((0,1)\), we can go up 2 units and then to the right 3 units to find another point on the line (for example, \((3,3)\)).

Step3: Shade the region

Since the inequality is \(y\geq\frac{2}{3}x + 1\), we shade the region above the boundary line. This is because for a given \(x\) - value, the \(y\) - values that satisfy \(y\geq\frac{2}{3}x + 1\) are the ones that are greater than or equal to the \(y\) - value on the line \(y = \frac{2}{3}x+1\).

(Note: Since the problem asks to graph the inequality, the final answer is the graph with a solid line \(y=\frac{2}{3}x + 1\) and the region above the line shaded. If we were to describe the steps for graphing, the above steps are the key parts of the process.)

Answer:

The graph has a solid line \(y = \frac{2}{3}x+1\) (passing through \((0,1)\) and \((3,3)\) etc.) and the region above the line is shaded.