Question

solve the system of two linear inequalities graphically.\

$$\begin{cases}3x + 8y < -24\\\\x \\geq 4\\end{cases}$$

\
step 1 of 3: graph the solution set of the first linear inequality.

Explanation:

Step1: Rewrite to slope-intercept form

Rearrange $3x + 8y < -24$ to solve for $y$:
$8y < -3x -24$
$y < -\frac{3}{8}x - 3$

Step2: Graph the boundary line

The boundary is the line $y = -\frac{3}{8}x - 3$. Since the inequality is $<$, draw a dashed line.

• Find intercepts:
• When $x=0$, $y=-3$ (y-intercept: $(0, -3)$)
• When $y=0$, $0 = -\frac{3}{8}x - 3 \implies x=-8$ (x-intercept: $(-8, 0)$)

Step3: Shade the solution region

Since $y < -\frac{3}{8}x - 3$, shade the area below the dashed boundary line.

Answer:

The graph of $3x + 8y < -24$ is a dashed line $y = -\frac{3}{8}x - 3$ (passing through $(-8,0)$ and $(0,-3)$) with the region below the line shaded.