Question

Two investments earn an annual income of $621. One investment earns an annual simple interest rate of 8.5%, and the other investment earns an annual simple interest rate of 6.1%. The total amount invested is $9000. How much is invested in each account?

Answer 1

\[x + y = 9000\]
\[.085x + .061y = 621\]
Solve system using substitution
\[y = 9000 - x\]
\[.085x + .061(9000-x) = 621\]
combine like terms
\[x(.085-.061) = 621 - 9000(.061)\]
\[x = \frac{72}{.024} = 3000\]
Therefore
\[y = 9000 - 3000 = 6000\]