Question

solve the compound inequality. use graphs to show the solution set to each of the two given inequalities, as well as a third graph that shows the solution set of the compound inequality. except for the empty set, express the solution set in interval notation.
\\(4(3 - x) < -2\\) and \\(\frac{x - 1}{4} \leq -1\\)

graph the solution set of \\(4(3 - x) < -2\\). choose the correct graph below.
\\(\bigcirc\\) a. \\(\bigcirc\\) b.
\\(\bigcirc\\) c. \\(\bigcirc\\) d.

graph the solution set of \\(\frac{x - 1}{4} \leq -1\\). choose the correct graph below.
\\(\bigcirc\\) a. \\(\bigcirc\\) b.
\\(\bigcirc\\) c. \\(\bigcirc\\) d.

Explanation:

Step1: Solve $4(3-x) < -2$

Expand and isolate $x$:
$12 - 4x < -2$
$-4x < -2 - 12$
$-4x < -14$
Divide by $-4$ (reverse inequality):
$x > \frac{14}{4} = 3.5$

Step2: Match to first graph

The solution $x > 3.5$ is an open circle at 3.5, shading right. This matches Option A.

Step3: Solve $\frac{x-1}{4} \leq -1$

Multiply by 4, isolate $x$:
$x - 1 \leq -4$
$x \leq -4 + 1$
$x \leq -3$

Step4: Match to second graph

The solution $x \leq -3$ is a closed circle at -3, shading left. This matches Option C.

Step5: Find compound solution

The compound inequality uses "and", so we need values that satisfy both $x > 3.5$ and $x \leq -3$. There is no overlap, so the solution is the empty set.

Answer:

1. Graph for $4(3-x) < -2$: A. (Open circle at 3.5, shading right)
2. Graph for $\frac{x-1}{4} \leq -1$: C. (Closed circle at -3, shading left)
3. Compound inequality solution: $\emptyset$ (empty set)