SOLUTION: annual interest on $18,000 investment exceeds interest earned on $13,000 investment by $558. The $18,000 is invested at a 6% higher interest rate than The $13,000 investment. Wha

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: annual interest on $18,000 investment exceeds interest earned on $13,000 investment by $558. The $18,000 is invested at a 6% higher interest rate than The $13,000 investment. Wha      Log On

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Question 659750: annual interest on $18,000 investment exceeds interest earned on $13,000 investment by $558. The $18,000 is invested at a 6% higher interest rate than The $13,000 investment. What is interest rate of each investment?
Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
x = interest on 18000
y = interest on 13000
18000x = 13000y + 558
x + .06 = y
Plug the value of y into the first equation
18000x = 13000(x + .06) + 558
Distribute the 13000 and combine like terms
18000x = 13000x + 780 + 558
18000x - 13000x = 13000x - 13000x + 1338
5000x = 1338
5000x/5000 = 1338/5000
x = .2676 or 27% for $18,000
Substitute x into second equation
.2676 + .06 = y = .3276 or 33% for $13,000
Proof
18000(.2676) = 4816.8
13000(.3276) = 4258.8
4816.8 - 4258.8 = 558