The problem requires graphing the solution to the following system of inequalities: $x - y \ge -6$ $-2x - y \le 5$

1. Problem Description

The problem requires graphing the solution to the following system of inequalities:

$x - y \ge -6$

$-2x - y \le 5$

2. Solution Steps

First, we rewrite the inequalities in slope-intercept form (i.e., $y = mx + b$).

For the first inequality, $x - y \ge -6$, we can rearrange it to isolate $y$:

$x + 6 \ge y$, or $y \le x + 6$

For the second inequality, $-2x - y \le 5$, we can rearrange it to isolate $y$:

$-y \le 2x + 5$

$y \ge -2x - 5$

Now we have the system of inequalities:

$y \le x + 6$

$y \ge -2x - 5$

To graph the solution, we need to graph the lines $y = x + 6$ and $y = -2x - 5$.

For the line $y = x + 6$, the y-intercept is 6 and the slope is

1. For the line $y = -2x - 5$, the y-intercept is -5 and the slope is -

2.

Since the first inequality is $y \le x + 6$, we shade the region below the line $y = x + 6$.

Since the second inequality is $y \ge -2x - 5$, we shade the region above the line $y = -2x - 5$.

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

The intersection point of the two lines can be found by setting $x + 6 = -2x - 5$, so $3x = -11$, and $x = -\frac{11}{3}$.

Then $y = -\frac{11}{3} + 6 = -\frac{11}{3} + \frac{18}{3} = \frac{7}{3}$.

So, the intersection point is $(-\frac{11}{3}, \frac{7}{3})$.

3. Final Answer

The solution to the system of inequalities $x - y \ge -6$ and $-2x - y \le 5$ is the region bounded by the lines $y = x + 6$ and $y = -2x - 5$, where $y \le x + 6$ and $y \ge -2x - 5$. The intersection point of these lines is $(-\frac{11}{3}, \frac{7}{3})$. We shade the region below $y=x+6$ and above $y=-2x-5$.