Question

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

$$\begin{cases} y < -5x - 10 \\\\ 4y \\geq 4x + 32 \\end{cases}$$

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

Explanation:

Step1: Simplify the second inequality

Divide all terms by 4:
$$\frac{4y}{4} \geq \frac{4x}{4} + \frac{32}{4}$$
$$y \geq x + 8$$

Step2: Graph boundary lines

1. For $y < -5x -10$: Draw a dashed line for $y=-5x-10$ (since the inequality is strict, no equality). The line has slope $-5$ and y-intercept $(0,-10)$.
2. For $y \geq x + 8$: Draw a solid line for $y=x+8$ (since the inequality includes equality). The line has slope $1$ and y-intercept $(0,8)$.

Step3: Find intersection of boundaries

Set $-5x -10 = x + 8$:
$$-5x - x = 8 + 10$$
$$-6x = 18$$
$$x = -3$$
Substitute $x=-3$ into $y=x+8$:
$$y = -3 + 8 = 5$$
Intersection point: $(-3, 5)$

Step4: Shade solution regions

1. For $y < -5x -10$: Shade below the dashed line (test point $(0,0)$: $0 < -10$ is false, so shade opposite side).
2. For $y \geq x + 8$: Shade above the solid line (test point $(0,0)$: $0 \geq 8$ is false, so shade opposite side).

Step5: Identify overlapping region

The solution set is the overlapping shaded area that satisfies both inequalities, bounded by the dashed line $y=-5x-10$ (not included) and solid line $y=x+8$ (included), meeting at $(-3,5)$.

Answer:

The solution set is the overlapping shaded region:

• Below the dashed line $y = -5x - 10$
• Above the solid line $y = x + 8$
• The region is bounded by the two lines intersecting at $(-3, 5)$, with the dashed line not included in the solution and the solid line included.