Question

graph the inequality.
2x+3y≤12

Explanation:

1. Boundary Line: The equation \(2x + 3y = 12\) is linear. We find its intercepts to plot the line. The solid line is used because the inequality is non - strict (\(\leq\)).
2. Test Point: Using the origin as a test point is convenient. Substituting into the inequality shows that the origin is in the solution set, so we shade the region containing the origin. This gives the graph of the inequality \(2x + 3y\leq12\).

Answer:

To graph \(2x + 3y \leq 12\), follow these steps:

Step 1: Graph the boundary line

First, rewrite the inequality as an equation to find the boundary line: \(2x + 3y = 12\).

• Find the x - intercept: Set \(y = 0\), then \(2x=12\), so \(x = 6\). The x - intercept is \((6,0)\).
• Find the y - intercept: Set \(x = 0\), then \(3y=12\), so \(y = 4\). The y - intercept is \((0,4)\).
• Since the inequality is \(\leq\), the boundary line \(2x + 3y = 12\) should be a solid line (because the points on the line are included in the solution set). Plot the points \((6,0)\) and \((0,4)\) and draw a solid line through them.

Step 2: Shade the solution region

To determine which side of the line to shade, pick a test point not on the line. The origin \((0,0)\) is a good choice.
Substitute \(x = 0\) and \(y = 0\) into the inequality \(2x+3y\leq12\):
\(2(0)+3(0)=0\leq12\), which is true.
Since the test point \((0,0)\) satisfies the inequality, shade the region that includes the origin (the region below the line \(2x + 3y = 12\)).