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? \( \frac{x+2}{2 x}=\frac{x^{2}-7 x+10}{4 x} \) \( \frac{x-1}{2 x}=\frac{x^{2}-7 x+10}{4 x} \) \( \frac{x+1}{2 x}=\frac{x^{2}-7 x+10}{4 x} \)

We start with the original equation: \[ \frac{1}{2}+\frac{1}{2x}=\frac{x^2-7x+10}{4x} \] Notice that the left-hand side contains two fractions. To combine them, we need a common denominator. The fractions \(\frac{1}{2}\) and \(\frac{1}{2x}\) have denominators \(2\) and \(2x\) respectively. The least common denominator is \(2x\). Rewrite \(\frac{1}{2}\) so that it has the denominator \(2x\): \[ \frac{1}{2}=\frac{x}{2x} \] Now, the sum of the fractions becomes: \[ \frac{x}{2x}+\frac{1}{2x}=\frac{x+1}{2x} \] Thus, the original equation can be rewritten as the following proportion: \[ \frac{x+1}{2x}=\frac{x^2-7x+10}{4x} \]