\begin{tabular}{l} Solve \( \frac{1}{2}+\frac{1}{2 x}=\frac{x^{2}-7 x+10}{4 x} \) by rewriting the equation as a proportion. \\ Which proportion is equivalent to the original \\ equation? \\ \( \qquad \begin{array}{ll}\frac{x+2}{2 x}=\frac{x^{2}-7 x+10}{4 x} & \text { Name the true solution(s) to the equation. } \\ \frac{x-1}{2 x}=\frac{x^{2}-7 x+10}{4 x} & x=0 \\ x=1 \\ 2 x & \frac{x^{2}-7 x+10}{4 x} \\ x=8 \\ x=0, x=1, \text { and } x=8 \\ x=1 \text { and } x=8 \\ \text { There are no solutions. }\end{array} \) \\ \hline\end{tabular}

To solve the equation \( \frac{1}{2}+\frac{1}{2 x}=\frac{x^{2}-7 x+10}{4 x} \), we can start rewriting it in a way that helps us find a proportion. By getting a common denominator on the left side, we can transform it into \( \frac{x+2}{2x} = \frac{x^2 - 7x + 10}{4x} \). Cross-multiplying, we end up with \( 4(x + 2) = 2(x^2 - 7x + 10) \). The next step is to solve the quadratic equation we formed by simplifying this proportion. After distributing and collecting like terms, we can factor the left side: \( 2x^2 - 6x = 0 \) breaks down to \( 2x(x - 3) = 0 \). This gives us two potential solutions: \( x = 0 \) and \( x = 3 \). However, substituting back into the original equation reveals that \( x = 0 \) is invalid due to division by zero! Therefore, the only valid solution is \( x = 3 \). Thus, the only true solution to the equation is \( x = 3 \). Happy solving!