Question

solve for q and graph the solution. 8 ≥ |q + 3| + 4 click two endpoints to graph a line segment, an endpo ray, or two arrowheads to graph a line. to change endpoints circles, click on them. -7 -6 -5 -4 -3 -2 -1 0

Explanation:

Step1: Isolate the absolute value

Subtract 4 from both sides of the inequality $8 \geq |q + 3| + 4$.
$8 - 4 \geq |q + 3| + 4 - 4$
$4 \geq |q + 3|$ which is equivalent to $|q + 3| \leq 4$.

Step2: Solve the absolute value inequality

The absolute value inequality $|q + 3| \leq 4$ means that $-4 \leq q + 3 \leq 4$.

Step3: Solve for q

Subtract 3 from all parts of the compound inequality.
For the left part: $-4 - 3 \leq q + 3 - 3$
$-7 \leq q$
For the right part: $q + 3 - 3 \leq 4 - 3$
$q \leq 1$
So the solution is $-7 \leq q \leq 1$.

Answer:

The solution for \( q \) is \( -7 \leq q \leq 1 \). To graph this, we would plot a line segment on the number line with endpoints at \( -7 \) (closed circle, since the inequality is "less than or equal to") and \( 1 \) (closed circle), and shade the region between them.