\( 1<\begin{array}{l}\text { Solve and check your solution. } \\ \qquad \begin{array}{l}\frac{1}{8} x+2=-8\end{array} \\ x=\square\end{array} \)

To solve the equation \(\frac{1}{8} x + 2 = -8\), start by isolating \(x\). First, subtract 2 from both sides: \[ \frac{1}{8} x + 2 - 2 = -8 - 2 \] This simplifies to: \[ \frac{1}{8} x = -10 \] Next, multiply both sides by 8 to eliminate the fraction: \[ x = -10 \times 8 = -80 \] Now, check your solution by substituting \(-80\) back into the original equation: \[ \frac{1}{8}(-80) + 2 = -8 \] Calculate \(\frac{1}{8}(-80) = -10\), so: \[ -10 + 2 = -8 \] Since both sides match, \(x = -80\) is indeed the correct solution! So the final answer is: \(x = -80\)