The problem asks to graph the solution to the following system of inequalities: $y \ge 3x-1$ $x+y \le 5$

1. Problem Description

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

$y \ge 3x-1$

$x+y \le 5$

2. Solution Steps

First, let's rewrite the second inequality:

$x+y \le 5$

$y \le -x+5$

Now, let's graph both inequalities.

For the first inequality, $y \ge 3x-1$:

The boundary line is $y=3x-1$.

When $x=0$, $y=-1$.

When $x=1$, $y=3(1)-1=2$.

So, the line passes through $(0, -1)$ and $(1, 2)$.

Since the inequality is $y \ge 3x-1$, we shade the region above the line. The line itself is included since it's "greater than or equal to."

For the second inequality, $y \le -x+5$:

The boundary line is $y=-x+5$.

When $x=0$, $y=5$.

When $x=5$, $y=0$.

So, the line passes through $(0, 5)$ and $(5, 0)$.

Since the inequality is $y \le -x+5$, we shade the region below the line. The line itself is included since it's "less than or equal to."

The solution to the system of inequalities is the region where the shaded regions of both inequalities overlap, including the boundary lines.

3. Final Answer

The solution is the intersection of the regions $y \ge 3x-1$ and $y \le -x+5$. To graph this, we would graph the lines $y = 3x-1$ and $y = -x+5$ and shade the region that satisfies both inequalities. The region is bounded by the lines $y=3x-1$ and $y=-x+5$.