Question

solve the system of two linear inequalities graphically.\
\\(\

$$\begin{cases} 2x + 8y < -8 \\\\ x \\geq 5 \\end{cases}$$

\\)\
step 3 of 3: graph the solution set for the system.

Explanation:

Step1: Rewrite first inequality to slope-intercept form

Isolate $y$ in $2x + 8y < -8$:

$$\begin{align*} 8y &< -2x -8 \\ y &< -\frac{2}{8}x - \frac{8}{8} \\ y &< -\frac{1}{4}x -1 \end{align*}$$

Step2: Identify boundary lines

• For $y < -\frac{1}{4}x -1$: Boundary is $y = -\frac{1}{4}x -1$ (dashed line, since inequality is strict $<$).
• For $x \geq 5$: Boundary is $x=5$ (solid vertical line, since inequality is $\geq$).

Step3: Determine shaded regions

• For $y < -\frac{1}{4}x -1$: Shade below the dashed boundary line.
• For $x \geq 5$: Shade to the right of the solid vertical line $x=5$.

Step4: Find overlapping shaded region

The solution set is the area that is shaded by both conditions: right of $x=5$ and below $y = -\frac{1}{4}x -1$.

Answer:

The solution set is the overlapping region:

1. Draw a dashed line for $y = -\frac{1}{4}x - 1$, shade the area below it.
2. Draw a solid vertical line for $x=5$, shade the area to its right.
3. The intersection of these two shaded areas is the solution (no points satisfy both conditions, so the solution set is empty, as substituting $x=5$ into the first inequality gives $y < -\frac{5}{4}-1=-\frac{9}{4}$, but the overlap exists as the region right of $x=5$ and below the dashed line; however, there is no real conflict, the graphical solution is the overlapping shaded area described).

Note: Upon checking, when $x\geq5$, $-\frac{1}{4}x -1 \leq -\frac{5}{4}-1=-\frac{9}{4}$, so the solution is all points where $x\geq5$ and $y < -\frac{1}{4}x -1$, represented by the overlapping shaded region.