The problem asks to graph the region determined by the following system of inequalities: $y \le 4x + 5$ $3y + 5x \le 12$ $x \ge -2$ $y \ge -5$

1. Problem Description

The problem asks to graph the region determined by the following system of inequalities:

$y \le 4x + 5$

$3y + 5x \le 12$

$x \ge -2$

$y \ge -5$

2. Solution Steps

First inequality: $y \le 4x + 5$.

This is a linear inequality. The boundary line is $y = 4x + 5$. The region is below this line.

Second inequality: $3y + 5x \le 12$.

Rearrange to get $3y \le -5x + 12$, so $y \le -\frac{5}{3}x + 4$. The boundary line is $y = -\frac{5}{3}x + 4$. The region is below this line.

Third inequality: $x \ge -2$.

This represents the region to the right of the vertical line $x = -2$.

Fourth inequality: $y \ge -5$.

This represents the region above the horizontal line $y = -5$.

We need to find the region that satisfies all four inequalities. The region is bounded by the lines:

$y = 4x + 5$

$y = -\frac{5}{3}x + 4$

$x = -2$

$y = -5$

The vertices of the feasible region are the intersection points of these lines.

Intersection of $x = -2$ and $y = -5$: $(-2, -5)$

Intersection of $x = -2$ and $y = 4x + 5$: $y = 4(-2) + 5 = -8 + 5 = -3$. So, $(-2, -3)$.

Intersection of $x = -2$ and $y = -\frac{5}{3}x + 4$: $y = -\frac{5}{3}(-2) + 4 = \frac{10}{3} + \frac{12}{3} = \frac{22}{3}$. So, $(-2, \frac{22}{3})$.

Intersection of $y = -5$ and $y = 4x + 5$: $-5 = 4x + 5$, so $4x = -10$, and $x = -\frac{10}{4} = -\frac{5}{2}$. So, $(-\frac{5}{2}, -5)$.

Intersection of $y = -5$ and $y = -\frac{5}{3}x + 4$: $-5 = -\frac{5}{3}x + 4$, so $-\frac{5}{3}x = -9$, and $x = \frac{27}{5}$. So, $(\frac{27}{5}, -5)$.

Intersection of $y = 4x + 5$ and $y = -\frac{5}{3}x + 4$: $4x + 5 = -\frac{5}{3}x + 4$.

$12x + 15 = -5x + 12$.

$17x = -3$.

$x = -\frac{3}{17}$.

$y = 4(-\frac{3}{17}) + 5 = -\frac{12}{17} + \frac{85}{17} = \frac{73}{17}$. So, $(-\frac{3}{17}, \frac{73}{17})$.

The feasible region is the region bounded by the inequalities.

3. Final Answer

The region is bounded by the lines:

$y \le 4x + 5$

$3y + 5x \le 12$

$x \ge -2$

$y \ge -5$

The solution is the graph of the region defined by these inequalities.