SOLUTION: A person has invested 6000 pesos. Part of the money is invested at 3% and the remainder at 4%. The annual income from the two interests is 225 pesos. How much is invested at each r

Algebra ->  Customizable Word Problem Solvers  -> Finance -> SOLUTION: A person has invested 6000 pesos. Part of the money is invested at 3% and the remainder at 4%. The annual income from the two interests is 225 pesos. How much is invested at each r      Log On

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Question 977565: A person has invested 6000 pesos. Part of the money is invested at 3% and the remainder at 4%. The annual income from the two interests is 225 pesos. How much is invested at each rate?
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
Total amount of money invested: $6000
x+y=6000,
Total yearly interest for the two accounts is: $225
0.03*x+0.04*y=225
x=6000-y
Substitute for x
0.03*(6000-y)+0.04*y=225
Multiply out
180-0.03*y+0.04*y=225
Combine like terms.
0.01*y=45
Isolate y
y=$ 4500.00 at 4% earns $180 interest
x=6000-y
Calculate x
x=$ 1500.00 at 3% earns $45 interest
Check
0.03*1500+0.04*4500=225
45+180=225
225=225
If 225=225 is TRUE and neither x nor y is negative then all is well
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