1. **State the problem:** Find the number of combinations of nickels (5 cents) and quarters (25 cents) that total 80 cents using exactly 11 coins. 2. **Define variables:** Let $n$ be the number of nickels and $q$ be the number of quarters. 3. **Write the system of equations:** $$n + q = 11$$ $$5n + 25q = 80$$ 4. **Solve the first equation for $n$:** $$n = 11 - q$$ 5. **Substitute $n$ into the second equation:** $$5(11 - q) + 25q = 80$$ 6. **Distribute and simplify:** $$55 - 5q + 25q = 80$$ $$55 + 20q = 80$$ 7. **Isolate $q$:** $$20q = 80 - 55$$ $$20q = 25$$ 8. **Divide both sides by 20:** $$q = \frac{\cancel{25}}{\cancel{20}}$$ Since 25 and 20 have no common factors other than 5, simplify: $$q = \frac{25}{20} = \frac{5}{4} = 1.25$$ 9. **Interpretation:** $q$ must be an integer number of coins, but $1.25$ is not an integer. So no solution here. 10. **Check integer values of $q$ from 0 to 11:** For each integer $q$, compute $n = 11 - q$ and check if $5n + 25q = 80$. - $q=0$: $n=11$, total = $5*11 + 25*0 = 55$ (not 80) - $q=1$: $n=10$, total = $5*10 + 25*1 = 50 + 25 = 75$ (not 80) - $q=2$: $n=9$, total = $5*9 + 25*2 = 45 + 50 = 95$ (not 80) - $q=3$: $n=8$, total = $5*8 + 25*3 = 40 + 75 = 115$ (not 80) - $q=4$: $n=7$, total = $5*7 + 25*4 = 35 + 100 = 135$ (not 80) - $q=5$: $n=6$, total = $5*6 + 25*5 = 30 + 125 = 155$ (not 80) - $q=6$: $n=5$, total = $5*5 + 25*6 = 25 + 150 = 175$ (not 80) - $q=7$: $n=4$, total = $5*4 + 25*7 = 20 + 175 = 195$ (not 80) - $q=8$: $n=3$, total = $5*3 + 25*8 = 15 + 200 = 215$ (not 80) - $q=9$: $n=2$, total = $5*2 + 25*9 = 10 + 225 = 235$ (not 80) - $q=10$: $n=1$, total = $5*1 + 25*10 = 5 + 250 = 255$ (not 80) - $q=11$: $n=0$, total = $5*0 + 25*11 = 0 + 275 = 275$ (not 80) 11. **Conclusion:** There are no integer combinations of nickels and quarters totaling 80 cents with exactly 11 coins. **Final answer:** No valid combinations exist.