Question: (\[ \frac{|x + 6|}{8} \leq 1 \] \[ x \leq \boxed{?} \] \[ x \geq \boxed{?} \]\)…

\(\[ \frac{|x + 6|}{8} \leq 1 \]

\[ x \leq \boxed{?} \]

\[ x \geq \boxed{?} \]\)

Solution

To solve the inequality \[ \frac{|x + 6|}{8} \leq 1 \] we first eliminate the fraction by multiplying both sides by 8: \[ |x + 6| \leq 8 \] This absolute value inequality implies two cases: Case 1: \[ x + 6 \leq 8 \] Subtract 6 from both sides: \[ x \leq 2 \] Case 2: \[ -(x + 6) \leq 8 \] Simplify the inequality: \[ -x - 6 \leq 8 \] Add 6 to both sides: \[ -x \leq 14 \] Multiply both sides by -1 (which reverses the inequality sign): \[ x \geq -14 \] Combining both cases, the solution is: \[ \begin{cases} x \leq 2 & \\ x \geq -14 & \end{cases} \] So, the values for \( x \) are: \(-14 \leq x \leq 2\)