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.

9514 1404 393

Step-by-step 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

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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%.

8(-10x+3.5y-7)

8×-10x + 8×3.5y + 8×-7

-80x + 28y - 56

therefore

-80x+28y-56 is the correct answer