We are asked to graph the region determined by the following system of inequalities: $y \le 3x + 5$ $6y + 5x \le 12$ $x \ge -2$ $y \ge -4$

1. Problem Description

We are asked to graph the region determined by the following system of inequalities:

$y \le 3x + 5$

$6y + 5x \le 12$

$x \ge -2$

$y \ge -4$

2. Solution Steps

First, we will rewrite the inequalities in slope-intercept form when applicable, or identify them as vertical or horizontal lines.

1. $y \le 3x + 5$ is already in slope-intercept form.

2. $6y + 5x \le 12$ can be rewritten as $6y \le -5x + 12$, which means $y \le -\frac{5}{6}x + 2$.

3. $x \ge -2$ is a vertical line.

4. $y \ge -4$ is a horizontal line.

Now we will graph the inequalities.

1. $y \le 3x + 5$ is the region below the line $y = 3x + 5$. The line has a y-intercept of 5 and a slope of

3.

2. $y \le -\frac{5}{6}x + 2$ is the region below the line $y = -\frac{5}{6}x + 2$. The line has a y-intercept of 2 and a slope of $-\frac{5}{6}$.

3. $x \ge -2$ is the region to the right of the vertical line $x = -2$.

4. $y \ge -4$ is the region above the horizontal line $y = -4$.

The solution to the system of inequalities is the region where all four inequalities are satisfied. It is the intersection of the four regions described above.

The region will be a closed polygon bounded by the lines $y = 3x+5$, $y = -\frac{5}{6}x + 2$, $x=-2$, and $y=-4$.

3. Final Answer

The solution is the region bounded by the inequalities $y \le 3x + 5$, $y \le -\frac{5}{6}x + 2$, $x \ge -2$, and $y \ge -4$. This region should be graphed using the provided graphing tool.