Question

solve for b.
7|b| + 7 ≤ 96
write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. use integers, fractions, or improper fractions in simplest form.

Explanation:

Step1: Subtract 7 from both sides

To isolate the absolute value term, we subtract 7 from both sides of the inequality \(7|b| + 7 \leq 96\).
\[7|b| + 7 - 7 \leq 96 - 7\]
\[7|b| \leq 89\]

Step2: Divide both sides by 7

Next, we divide both sides by 7 to solve for \(|b|\).
\[\frac{7|b|}{7} \leq \frac{89}{7}\]
\[|b| \leq \frac{89}{7}\]

Step3: Solve the absolute value inequality

The absolute value inequality \(|b| \leq \frac{89}{7}\) is equivalent to the compound inequality \(-\frac{89}{7} \leq b \leq \frac{89}{7}\) because if \(|x| \leq a\) (where \(a \geq 0\)), then \(-a \leq x \leq a\).

Answer:

\(-\frac{89}{7} \leq b \leq \frac{89}{7}\)