SOLUTION: $500,000 was borrowed in 3 loans, at 7%, 8%, and 9%. The annual interest was $38,600. If 3 times as much was borrowed at 7% than at 9%, how much was borrowed at each rate?

Algebra ->  Linear-equations -> SOLUTION: $500,000 was borrowed in 3 loans, at 7%, 8%, and 9%. The annual interest was $38,600. If 3 times as much was borrowed at 7% than at 9%, how much was borrowed at each rate?      Log On


   

Question 1115540: $500,000 was borrowed in 3 loans, at 7%, 8%, and 9%. The annual interest was $38,600. If 3 times as much was borrowed at 7% than at 9%, how much was borrowed at each rate?
Answer by addingup(3677) About Me  (Show Source):
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0.07x + 0.09y + 0.08(500,000-x+y) = 38,600
x = 3y
0.07(3y) + 0.09y + 0.08(500,000-4y) = 38600
0.21y + 0.09y + 40,000 - 0.32y = 38600
-0.02y = -1400
y = 70,000 amount at 9%
70,000(3) = 210,000 amount at 7%
500,000 - 280,000 = 220,000 amount at 8%
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Check:
70,000(0.09) + 210,000(0.07) + 220,000(0.08) = 38,600 Correct