Question

solve the system of two linear inequalities graphically.\

$$\begin{cases} 2y < -x + 2 \\\\ 2y \\geq 6x - 12 \\end{cases}$$

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

Explanation:

Step1: Rewrite inequalities to slope-intercept form

First inequality: $2y < -x + 2$ → $y < -\frac{1}{2}x + 1$
Second inequality: $2y \geq 6x - 12$ → $y \geq 3x - 6$

Step2: Graph boundary lines

For $y = -\frac{1}{2}x + 1$: use a dashed line (since inequality is $<$), y-intercept at $(0,1)$, slope $-\frac{1}{2}$.
For $y = 3x - 6$: use a solid line (since inequality is $\geq$), y-intercept at $(0,-6)$, slope $3$.

Step3: Shade solution regions

For $y < -\frac{1}{2}x + 1$: shade below the dashed line.
For $y \geq 3x - 6$: shade above the solid line.

Step4: Identify overlapping shaded area

The solution set is the intersection of the two shaded regions.

Answer:

The solution is the overlapping shaded area where:

1. Below the dashed line $y = -\frac{1}{2}x + 1$
2. Above the solid line $y = 3x - 6$

The intersection point of the two boundaries is found by solving $-\frac{1}{2}x + 1 = 3x - 6$:
$-\frac{1}{2}x - 3x = -6 - 1$
$-\frac{7}{2}x = -7$
$x = 2$, substitute back to get $y = 0$. So the lines intersect at $(2, 0)$, and the solution set is the region bounded by this point, the dashed line, and the solid line.