1. **State the problem:** We are given two equations involving $a$, $b$, and their product $ab$: $$\frac{ab}{3a + 2b} = \frac{1}{5}$$ and $$\frac{ab}{5a + 7b} = 1$$ We need to find the value of: $$\frac{ab}{a + b}$$ 2. **Write down the given equations clearly:** From the first equation: $$\frac{ab}{3a + 2b} = \frac{1}{5} \implies ab = \frac{3a + 2b}{5}$$ From the second equation: $$\frac{ab}{5a + 7b} = 1 \implies ab = 5a + 7b$$ 3. **Set the two expressions for $ab$ equal to each other:** $$\frac{3a + 2b}{5} = 5a + 7b$$ Multiply both sides by 5 to clear the denominator: $$3a + 2b = 5(5a + 7b) = 25a + 35b$$ 4. **Rearrange to isolate terms:** $$3a + 2b = 25a + 35b$$ Move all terms to one side: $$3a + 2b - 25a - 35b = 0$$ Simplify: $$-22a - 33b = 0$$ 5. **Divide entire equation by -11 to simplify:** $$\cancel{-11} \times 2a + \cancel{-11} \times 3b = 0 \implies 2a + 3b = 0$$ 6. **Express $a$ in terms of $b$:** $$2a = -3b \implies a = -\frac{3}{2}b$$ 7. **Substitute $a$ into the expression for $ab$ from the second equation:** Recall: $$ab = 5a + 7b$$ Substitute $a = -\frac{3}{2}b$: $$ab = 5\left(-\frac{3}{2}b\right) + 7b = -\frac{15}{2}b + 7b = \left(-\frac{15}{2} + 7\right)b = \left(-\frac{15}{2} + \frac{14}{2}\right)b = -\frac{1}{2}b$$ 8. **Calculate $ab$ explicitly:** Since $a = -\frac{3}{2}b$, then $$ab = a \times b = \left(-\frac{3}{2}b\right) \times b = -\frac{3}{2}b^2$$ From step 7, we also have: $$ab = -\frac{1}{2}b$$ Set equal: $$-\frac{3}{2}b^2 = -\frac{1}{2}b$$ Multiply both sides by 2 to clear denominators: $$-3b^2 = -b$$ Divide both sides by $-b$ (assuming $b \neq 0$): $$\cancel{-b} \times 3b = \cancel{-b} \times 1 \implies 3b = 1$$ 9. **Solve for $b$:** $$b = \frac{1}{3}$$ 10. **Find $a$ using $a = -\frac{3}{2}b$:** $$a = -\frac{3}{2} \times \frac{1}{3} = -\frac{1}{2}$$ 11. **Calculate $ab$:** $$ab = a \times b = -\frac{1}{2} \times \frac{1}{3} = -\frac{1}{6}$$ 12. **Calculate $a + b$:** $$a + b = -\frac{1}{2} + \frac{1}{3} = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6}$$ 13. **Calculate the desired value:** $$\frac{ab}{a + b} = \frac{-\frac{1}{6}}{-\frac{1}{6}} = 1$$ **Final answer:** $$\boxed{1}$$