1. **State the problem:** Solve for $z$ in the equation $$\frac{5}{5x+7} + \frac{z}{2x+7} = \frac{1}{7-2x}.$$ 2. **Identify the goal:** We want to isolate $z$ and express it in terms of $x$. 3. **Rewrite the equation:** $$\frac{5}{5x+7} + \frac{z}{2x+7} = \frac{1}{7-2x}.$$ 4. **Isolate the term with $z$:** Subtract $\frac{5}{5x+7}$ from both sides: $$\frac{z}{2x+7} = \frac{1}{7-2x} - \frac{5}{5x+7}.$$ 5. **Find a common denominator on the right side:** The denominators are $(7-2x)$ and $(5x+7)$. The common denominator is $(7-2x)(5x+7)$. Rewrite each fraction: $$\frac{1}{7-2x} = \frac{5x+7}{(7-2x)(5x+7)},$$ $$\frac{5}{5x+7} = \frac{5(7-2x)}{(5x+7)(7-2x)}.$$ 6. **Subtract the fractions:** $$\frac{1}{7-2x} - \frac{5}{5x+7} = \frac{5x+7 - 5(7-2x)}{(7-2x)(5x+7)}.$$ 7. **Simplify the numerator:** Calculate $5(7-2x) = 35 - 10x$, so numerator is: $$5x + 7 - (35 - 10x) = 5x + 7 - 35 + 10x = (5x + 10x) + (7 - 35) = 15x - 28.$$ 8. **Rewrite the equation for $z$:** $$\frac{z}{2x+7} = \frac{15x - 28}{(7-2x)(5x+7)}.$$ 9. **Multiply both sides by $(2x+7)$ to solve for $z$:** $$z = \frac{15x - 28}{(7-2x)(5x+7)} \times (2x+7).$$ 10. **Final expression for $z$:** $$z = \frac{(15x - 28)(2x + 7)}{(7 - 2x)(5x + 7)}.$$ This is the solution for $z$ in terms of $x$.