Question

solve the system of two linear inequalities graphically.\

$$\begin{cases}y \\leq -5x - 10 \\\\ y > x + 2\\end{cases}$$

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

Explanation:

Step1: Find intersection of lines

Set $-5x - 10 = x + 2$
$-6x = 12$
$x = -2$
Substitute $x=-2$ into $y=x+2$: $y = 0$
Intersection point: $(-2, 0)$

Step2: Graph boundary lines

1. For $y \leq -5x - 10$:

• Draw solid line $y = -5x - 10$ (since $\leq$ includes equality).
• Shade below the line (test point $(0,0)$: $0 \leq -10$ is false, so shade opposite side).

1. For $y > x + 2$:

• Draw dashed line $y = x + 2$ (since $>$ excludes equality).
• Shade above the line (test point $(0,0)$: $0 > 2$ is false, so shade opposite side).

Step3: Identify solution region

The solution set is the overlapping shaded area, which is the region that is above the dashed line $y=x+2$ and below the solid line $y=-5x-10$, bounded on the left by their intersection at $(-2, 0)$.

Answer:

The solution is the overlapping shaded region: above the dashed line $y = x + 2$, below the solid line $y = -5x - 10$, with the boundary intersection at $(-2, 0)$.
(Graph description: Solid line $y=-5x-10$ passing through $(0,-10)$ and $(-2,0)$; dashed line $y=x+2$ passing through $(0,2)$ and $(-2,0)$; overlapping shade is the area that satisfies both inequalities.)