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question graph the inequality on the axes below. 2x + 3y ≤ 3

Question

question graph the inequality on the axes below. 2x + 3y ≤ 3

Explanation:

Step1: Rewrite in slope - intercept form

First, rewrite $2x + 3y\leq3$ as $y\leq-\frac{2}{3}x + 1$.

Step2: Find the boundary line

The boundary line is $y =-\frac{2}{3}x + 1$. When $x = 0$, $y=1$; when $y = 0$, $x=\frac{3}{2}=1.5$. Plot the points $(0,1)$ and $(1.5,0)$ and draw a solid line (since the inequality is $\leq$).

Step3: Test a point

Test the point $(0,0)$. Substitute $x = 0$ and $y = 0$ into the original inequality $2(0)+3(0)\leq3$, which is true. So, shade the region that contains the origin.

Answer:

Graph a solid line through $(0,1)$ and $(1.5,0)$ and shade the region below the line (including the line) that contains the origin.