Marco has $\$8{,}000$ to invest. He puts part of it in an account earning $3\%$ interest and the rest in an account earning $8\%$. After one year, the total interest earned is $\$500$. How much did Marco put in the $3\%$ account?

Let $x$ be the amount invested at $3\%$. Then the amount at $8\%$ is $8{,}000 - x$. Total interest: $0.03 x + 0.08 (8{,}000 - x) = 500$. Distribute: $0.03 x + 640 - 0.08 x = 500$. Combine: $-0.05 x + 640 = 500$. Subtract $640$: $-0.05 x = -140$. Divide: $x = 2{,}800$. So $\$2{,}800$ is at $3\%$ (and $\$5{,}200$ is at $8\%$). Check: $0.03(2{,}800) + 0.08(5{,}200) = 84 + 416 = \$500$ ✓. - Trap D ($\$5{,}000$) reverses which account is which. - Trap A and C miss the algebra. The mixture model applies identically to coins and to investments: replace *coin value* with *interest rate* and the same equation structure does the job.