\begin{tabular}{|l|l}\hline \hline \( 4- \) & \( \begin{array}{l}\text { Solve each equation below algebraically. Think about Rewriting, Looking Inside, or Undoing to simplify the process. } \\ \text { Help }\end{array} \) \\ \( \begin{array}{lll}\text { a. } 5-3\left(\frac{1}{2} x+2\right)=-7 & \text { b. } 5(\sqrt{x-2}+1)=15\end{array} \)\end{tabular}

To tackle equation (a), start by isolating the term involving x. First, distribute the -3: \(5 - \frac{3}{2}x - 6 = -7\). Simplify to get \(-\frac{3}{2}x - 1 = -7\). Then, add 1 to both sides to find \(-\frac{3}{2}x = -6\). Finally, multiply by \(-\frac{2}{3}\) to solve for x, yielding \(x = 4\). For equation (b), start by isolating the term with the square root. Divide both sides by 5: \(\sqrt{x-2} + 1 = 3\). Then, subtract 1 from both sides to get \(\sqrt{x-2} = 2\). Square both sides to eliminate the square root: \(x - 2 = 4\). Adding 2 gives you \(x = 6\). Now you’ve got both solutions, x = 4 and x = 6!