Question

directions: graph each system of linear inequalities. cle

1. $y \leq -\frac{2}{3}x - 1$

$y \geq \frac{1}{2}x - 8$

1. $y$

Explanation:

Step1: Identify boundary lines

First, we take the inequalities as equations to find the boundary lines:

1. $y = -\frac{2}{3}x - 1$ (solid line, since $\leq$)
2. $y = \frac{1}{2}x - 8$ (solid line, since $\geq$)

Step2: Find intercepts for Line 1

For $y = -\frac{2}{3}x - 1$:

• x-intercept: Set $y=0$, $0 = -\frac{2}{3}x - 1 \implies x = -\frac{3}{2} = -1.5$
• y-intercept: Set $x=0$, $y = -1$

Step3: Find intercepts for Line 2

For $y = \frac{1}{2}x - 8$:

• x-intercept: Set $y=0$, $0 = \frac{1}{2}x - 8 \implies x = 16$
• y-intercept: Set $x=0$, $y = -8$

Step4: Find intersection of boundaries

Solve the system:

$$\begin{cases} y = -\frac{2}{3}x - 1 \\ y = \frac{1}{2}x - 8 \end{cases}$$

Set equal: $-\frac{2}{3}x - 1 = \frac{1}{2}x - 8$
Multiply by 6 to eliminate denominators: $-4x - 6 = 3x - 48$
Solve for $x$: $-7x = -42 \implies x=6$
Substitute $x=6$: $y = -\frac{2}{3}(6) -1 = -5$
Intersection point: $(6, -5)$

Step5: Test regions for inequalities

1. For $y \leq -\frac{2}{3}x -1$: Test $(0,0)$: $0 \leq -1$ is false, so shade below this line.
2. For $y \geq \frac{1}{2}x -8$: Test $(0,0)$: $0 \geq -8$ is true, so shade above this line.

Answer:

The solution is the overlapping shaded region, bounded by:

1. The solid line $y = -\frac{2}{3}x -1$ (shaded below it)
2. The solid line $y = \frac{1}{2}x -8$ (shaded above it)

with the intersection point of the two lines at $(6, -5)$. This overlapping region is the set of all points that satisfy both inequalities simultaneously.