1. **State the problem:** You bike 5 miles to campus and 5 miles back home, totaling 10 miles. You go faster going to campus (downhill) than returning by 9 miles per hour. The total time for the round trip is 1 hour 10 minutes ($= \frac{7}{6}$ hours). We need to find the average velocity on the return trip. 2. **Define variables:** Let $x$ be the speed (mph) on the return trip. Then the speed going to campus is $x + 9$ mph. 3. **Write time expressions:** Time going to campus = distance / speed = $\frac{5}{x+9}$ hours Time returning home = $\frac{5}{x}$ hours 4. **Write total time equation:** $$\frac{5}{x+9} + \frac{5}{x} = \frac{7}{6}$$ 5. **Multiply both sides by $x(x+9)$ to clear denominators:** $$5x + 5(x+9) = \frac{7}{6} x(x+9)$$ Simplify left side: $$5x + 5x + 45 = 10x + 45$$ 6. **Rewrite equation:** $$10x + 45 = \frac{7}{6} (x^2 + 9x)$$ Multiply both sides by 6: $$6(10x + 45) = 7(x^2 + 9x)$$ $$60x + 270 = 7x^2 + 63x$$ 7. **Bring all terms to one side:** $$7x^2 + 63x - 60x - 270 = 0$$ $$7x^2 + 3x - 270 = 0$$ 8. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=7$, $b=3$, and $c=-270$. Calculate discriminant: $$\Delta = 3^2 - 4\times7\times(-270) = 9 + 7560 = 7569$$ $$\sqrt{7569} = 87$$ 9. **Compute roots:** $$x = \frac{-3 \pm 87}{14}$$ Two solutions: $$x_1 = \frac{84}{14} = 6$$ $$x_2 = \frac{-90}{14} = -\frac{45}{7} \approx -6.43$$ Since speed cannot be negative, we take $x=6$ mph. 10. **Final answer:** The average velocity on the return trip is $\boxed{6}$ miles per hour.