Question

graph each system of linear inequalities. describe the

1. \

$$\begin{cases} y \\leq 3x + 6 \\\\ y < 3x - 8 \\end{cases}$$

\

$$\begin{cases} y \\geq \\frac{5}{4}x - 6 \\\\ y \\geq \\frac{5}{4}x \\end{cases}$$

Explanation:

Step1: Analyze first system's lines

The two inequalities are $y \leq 3x + 6$ and $y \leq 3x - 8$. Both have the same slope $m=3$, so they are parallel. The first line has a y-intercept of $(0,6)$ (solid line, since $\leq$ includes equality), and the second has a y-intercept of $(0,-8)$ (solid line, since $\leq$ includes equality).

Step2: Identify solution region for first system

For $y \leq 3x + 6$, shade below the line $y=3x+6$. For $y \leq 3x - 8$, shade below the line $y=3x-8$. Since $3x-8$ is always below $3x+6$ for all $x$, there is no overlap between the two shaded regions.

Step3: Analyze second system's lines

The two inequalities are $y \geq \frac{5}{4}x - 6$ and $y \geq \frac{5}{4}x$. Both have the same slope $m=\frac{5}{4}$, so they are parallel. The first line has a y-intercept of $(0,-6)$ (solid line, since $\geq$ includes equality), and the second has a y-intercept of $(0,0)$ (solid line, since $\geq$ includes equality).

Step4: Identify solution region for second system

For $y \geq \frac{5}{4}x - 6$, shade above the line $y=\frac{5}{4}x-6$. For $y \geq \frac{5}{4}x$, shade above the line $y=\frac{5}{4}x$. Since $y=\frac{5}{4}x$ is always above $y=\frac{5}{4}x-6$ for all $x$, the overlapping shaded region is the area above $y=\frac{5}{4}x$.

Answer:

1. For the system

$$\begin{cases} y \leq 3x + 6 \\ y \leq 3x - 8 \end{cases}$$

:
The two lines are parallel, and there is no overlapping solution region (no solution exists for the system).

1. For the system

$$\begin{cases} y \geq \frac{5}{4}x - 6 \\ y \geq \frac{5}{4}x \end{cases}$$

:
The two lines are parallel, and the solution region is all points above (and on) the line $y=\frac{5}{4}x$.