1. **State the problem:** We are given a system of equations involving dimes ($x$) and quarters ($y$): $$0.10x + 0.25y = 3.65$$ and a second equation that appears incomplete or incorrectly written, but from the later lines, we have two linear equations: $$50x + 30y = 550$$ $$44x + 36y = 532$$ We want to solve for $x$ and $y$. 2. **Rewrite the system clearly:** $$50x + 30y = 550$$ $$44x + 36y = 532$$ 3. **Use the elimination method:** Multiply the first equation by 6 and the second by 5 to align coefficients of $y$: $$6(50x + 30y) = 6(550) \Rightarrow 300x + 180y = 3300$$ $$5(44x + 36y) = 5(532) \Rightarrow 220x + 180y = 2660$$ 4. **Subtract the second from the first:** $$ (300x + 180y) - (220x + 180y) = 3300 - 2660 $$ $$ 80x + \cancel{180y} - \cancel{180y} = 640 $$ $$ 80x = 640 $$ 5. **Solve for $x$:** $$ x = \frac{640}{80} = 8 $$ 6. **Substitute $x=8$ into one original equation to find $y$:** Using $50x + 30y = 550$: $$ 50(8) + 30y = 550 $$ $$ 400 + 30y = 550 $$ $$ 30y = 550 - 400 = 150 $$ 7. **Solve for $y$:** $$ y = \frac{150}{30} = 5 $$ **Final answer:** $$ x = 8, \quad y = 5 $$ This means there are 8 dimes and 5 quarters.