graph the solution to this system of inequalities in the coordinate plane.\n3y > 2x + 12\n2x + y ≤ - 5

Explanation:

Step1: Rewrite inequalities in slope - intercept form

For $3y>2x + 12$, divide by 3 to get $y>\frac{2}{3}x+4$. For $2x + y\leq - 5$, rewrite as $y\leq - 2x-5$.

Step2: Graph $y=\frac{2}{3}x + 4$

Since $y>\frac{2}{3}x+4$, draw a dashed line (because the inequality is strict). The y - intercept is 4 and the slope is $\frac{2}{3}$.

Step3: Graph $y=-2x - 5$

Since $y\leq - 2x-5$, draw a solid line (because the inequality is non - strict). The y - intercept is - 5 and the slope is - 2.

Step4: Determine the shaded region

For $y>\frac{2}{3}x+4$, shade above the line. For $y\leq - 2x-5$, shade below the line. The solution of the system is the intersection of the two shaded regions.

Answer:

Graph a dashed line $y=\frac{2}{3}x + 4$, a solid line $y=-2x - 5$, shade above the first line and below the second line, and the overlapping shaded region is the solution.