We need to solve the system of linear inequalities: $y \ge -2x$ $y \ge 3x + 4$

1. Problem Description

We need to solve the system of linear inequalities:

$y \ge -2x$

$y \ge 3x + 4$

2. Solution Steps

To solve the system of inequalities, we can look for the region where both inequalities are satisfied.

First, we can find the intersection point of the two lines $y = -2x$ and $y = 3x + 4$.

Set $-2x = 3x + 4$.

Subtracting $3x$ from both sides gives $-5x = 4$.

Dividing both sides by $-5$ gives $x = -\frac{4}{5}$.

Now, substitute $x = -\frac{4}{5}$ into $y = -2x$ to find the corresponding $y$ value:

$y = -2(-\frac{4}{5}) = \frac{8}{5}$.

The intersection point is $(-\frac{4}{5}, \frac{8}{5})$.

To find the solution region, we can test a point to see which side of each line is the solution.

For the first inequality $y \ge -2x$, let's test the point $(0, 1)$. Since $1 \ge -2(0)$ or $1 \ge 0$ is true, the region above the line $y = -2x$ is the solution.

For the second inequality $y \ge 3x + 4$, let's test the point $(0, 5)$. Since $5 \ge 3(0) + 4$ or $5 \ge 4$ is true, the region above the line $y = 3x + 4$ is the solution.

The solution to the system of inequalities is the region where both inequalities are satisfied, i.e., the region above both lines $y = -2x$ and $y = 3x + 4$.

3. Final Answer

The solution to the system of inequalities is the region where $y \ge -2x$ and $y \ge 3x + 4$ are both true. This is the region above both lines $y = -2x$ and $y = 3x + 4$, which meet at the point $(-\frac{4}{5}, \frac{8}{5})$.