We are asked to graph the solution set of the following system of inequalities:

1. $4x + 8y \leq 8$
2. $2x + y \leq 4$

Step 1: Simplify the inequalities

First inequality:

$4x + 8y \leq 8$ Divide the entire inequality by 4: $x + 2y \leq 2$ This can be rewritten as: $2y \leq -x + 2$ $y \leq -\frac{1}{2}x + 1$

Second inequality:

$2x + y \leq 4$ This inequality is already simplified. You can rewrite it as: $y \leq -2x + 4$

Step 2: Graphing the system

Now, we will graph both inequalities:

• For $y \leq -\frac{1}{2}x + 1$, the boundary line is $y = -\frac{1}{2}x + 1$, with a slope of -1/2 and a y-intercept of 1.
• For $y \leq -2x + 4$, the boundary line is $y = -2x + 4$, with a slope of -2 and a y-intercept of 4.

Both lines will be solid since the inequalities are "less than or equal to" ($\leq$). The solution set is the region where the shaded areas of both inequalities overlap.

Step 3: Test a point to determine shading

• Use a test point, like (0,0), to see if it satisfies both inequalities.

For $y \leq -\frac{1}{2}x + 1$: $0 \leq -\frac{1}{2}(0) + 1 \implies 0 \leq 1 \quad (\text{True})$ For $y \leq -2x + 4$: $0 \leq -2(0) + 4 \implies 0 \leq 4 \quad (\text{True})$ Since (0,0) satisfies both inequalities, the shaded region for both will include the area containing the origin (0,0).

The solution region is where the shaded regions overlap.

Would you like further details on how to draw this graph or test more points?

Here are 5 related questions to explore:

1. How do we determine which region to shade for a system of inequalities?
2. What role does the slope play in graphing these inequalities?
3. How does changing the inequality sign affect the graph?
4. What are alternative methods to graph systems of inequalities?
5. How can we find the intersection points of these boundary lines algebraically?

Tip: Always choose a simple point like (0, 0) to check which side of the boundary line to shade unless the line passes through the origin, in which case you can select another point.