The problem asks us to graph the solution to the following system of inequalities: $y \ge -2x$ $y \ge 3x + 2$

1. Problem Description

The problem asks us to graph the solution to the following system of inequalities:

$y \ge -2x$

$y \ge 3x + 2$

2. Solution Steps

First, we graph the line $y = -2x$. Since the inequality is $y \ge -2x$, we shade the region above the line.

The line $y=-2x$ passes through the origin $(0,0)$. When $x=1$, $y=-2$. Thus, the line also passes through $(1, -2)$.

Since the inequality is $\ge$, the line is solid.

Next, we graph the line $y = 3x + 2$. Since the inequality is $y \ge 3x + 2$, we shade the region above the line.

The $y$-intercept is $(0, 2)$. When $x = -1$, $y = 3(-1) + 2 = -1$. Thus, the line also passes through $(-1, -1)$.

Since the inequality is $\ge$, the line is solid.

The solution to the system of inequalities is the region where the shaded regions of both inequalities overlap.

To find the intersection point of the two lines, we set $-2x = 3x + 2$.

$-5x = 2$

$x = -\frac{2}{5}$

$y = -2(-\frac{2}{5}) = \frac{4}{5}$

The intersection point is $(-\frac{2}{5}, \frac{4}{5})$.

3. Final Answer

The solution is the region above both lines $y=-2x$ and $y=3x+2$. The lines intersect at $(-\frac{2}{5}, \frac{4}{5})$. The lines are solid.