Question

Jarvis invested some money at 6% interest. Jarvis also invested $58 more than 3 times that amount at 9%. How much is invested at each rate if Jarvis receives $1097.19 in interest after one year? (Round to two decimal places if necessary.)

Use the variables x and y to set up a system of equations to solve the given problem.

Answer 1

9514 1404 393

Answer:

• $3309 at 6%
• $9985 at 9%

Explanation:

Let x and y represent amounts invested at 6% and 9%, respectively.

y = 3x +58 . . . . . . . the amount invested at 9%

0.06x +0.09y = 1097.19 . . . . . . total interest earned

__

Substituting for y, we have ...

0.06x +0.09(3x +58) = 1097.19

0.33x + 5.22 = 1097.19 . . . . . . . . . simplify

0.33x = 1091.97 . . . . . . . . . . . . subtract 5.22

x = 3309 . . . . . . . . . . . . . . . . divide by 0.33

y = 3(3309) +58 = 9985

$3309 is invested at 6%; $9985 is invested at 9%.