The problem requires us to graph the solution to the following system of inequalities: $y \ge 4x - 3$ $x + y \le 7$

1. Problem Description

The problem requires us to graph the solution to the following system of inequalities:

$y \ge 4x - 3$

$x + y \le 7$

2. Solution Steps

First, we need to rewrite the inequalities in slope-intercept form if necessary. The first inequality is already in slope-intercept form.

The second inequality can be rewritten as:

$y \le -x + 7$

Now we analyze each inequality.

For $y \ge 4x - 3$:

This is a linear inequality. We first consider the equation $y = 4x - 3$. The slope is 4 and the y-intercept is -

3. Since the inequality is $\ge$, we draw a solid line (because the points on the line are included in the solution). The region to be shaded is above the line.

For $y \le -x + 7$:

This is also a linear inequality. We first consider the equation $y = -x + 7$. The slope is -1 and the y-intercept is

7. Since the inequality is $\le$, we draw a solid line (because the points on the line are included in the solution). The region to be shaded is below the line.

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

To accurately graph these, plot at least two points for each line:

For $y = 4x - 3$:

If $x=0$, $y = -3$. Point (0, -3).

If $x=1$, $y = 4(1) - 3 = 1$. Point (1, 1).

For $y = -x + 7$:

If $x=0$, $y = 7$. Point (0, 7).

If $x=7$, $y = 0$. Point (7, 0).

Graph these lines, and shade the appropriate regions based on the inequalities.

3. Final Answer

The final answer is the graph of the solution region obtained by graphing the two linear inequalities: $y \ge 4x-3$ and $x+y \le 7$. The region is bounded by the lines $y = 4x-3$ and $y = -x+7$. The solution area includes the lines themselves, is above $y=4x-3$, and below $y = -x+7$. The intersection point can be found by setting $4x-3 = -x+7$, so $5x = 10$, and $x=2$. Then $y = -2+7 = 5$. The lines intersect at (2, 5).