1. **State the problem:** Solve the equation $$\frac{5}{2w - 3} - \frac{7}{7w} = 1$$ for $w$. 2. **Identify the common denominator:** The denominators are $2w - 3$ and $7w$. The common denominator is $$7w(2w - 3)$$. 3. **Multiply both sides by the common denominator to clear fractions:** $$7w(2w - 3) \times \left(\frac{5}{2w - 3} - \frac{7}{7w}\right) = 7w(2w - 3) \times 1$$ 4. **Distribute and simplify:** $$7w \cancel{(2w - 3)} \times \frac{5}{\cancel{2w - 3}} - (2w - 3) \cancel{7w} \times \frac{7}{\cancel{7w}} = 7w(2w - 3)$$ which simplifies to $$7w \times 5 - (2w - 3) \times 7 = 7w(2w - 3)$$ 5. **Expand both sides:** $$35w - 7(2w - 3) = 7w(2w - 3)$$ $$35w - 14w + 21 = 14w^2 - 21w$$ 6. **Combine like terms on the left:** $$21w + 21 = 14w^2 - 21w$$ 7. **Bring all terms to one side to set equation to zero:** $$0 = 14w^2 - 21w - 21w - 21$$ $$0 = 14w^2 - 42w - 21$$ 8. **Divide entire equation by 7 to simplify:** $$0 = 2w^2 - 6w - 3$$ 9. **Use the quadratic formula to solve for $w$:** $$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-6$, $c=-3$. 10. **Calculate the discriminant:** $$b^2 - 4ac = (-6)^2 - 4 \times 2 \times (-3) = 36 + 24 = 60$$ 11. **Find the roots:** $$w = \frac{6 \pm \sqrt{60}}{4} = \frac{6 \pm 2\sqrt{15}}{4} = \frac{3 \pm \sqrt{15}}{2}$$ 12. **Check for restrictions:** The denominators $2w - 3$ and $7w$ cannot be zero. - For $2w - 3 = 0$, $w = \frac{3}{2}$. - For $7w = 0$, $w = 0$. Neither $\frac{3 + \sqrt{15}}{2}$ nor $\frac{3 - \sqrt{15}}{2}$ equals $\frac{3}{2}$ or $0$, so both solutions are valid. **Final answer:** $$w = \frac{3 + \sqrt{15}}{2} \quad \text{or} \quad w = \frac{3 - \sqrt{15}}{2}$$