SOLUTION: A sum of $37,500 is invested, part of it at 11% annual interest and the remainder at 8% annual interest. If the yearly interest from the 11% investment is $303 less than twice the

Question 160209: A sum of $37,500 is invested, part of it at 11% annual interest and the remainder at 8% annual interest. If the yearly interest from the 11% investment is $303 less than twice the interest from the 8% investment, how much is invested at each interest rate?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.11x=2[.08(37,500-x)]-303
.11x=2[(3000-.08x)]-303
.11x=6000-.16x-303
.11x+.16x=5697
.27x=5697
x=5697/.27
x=21,100 invested @ 11%.
37,500-21,100=16,400 invested @ 8%
Proof:
.11*21,100=2(.08*16,400)-303
2,321=2*1,312-303
2,321=2,624-303
2,321=2,321
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