SOLUTION: I would really appreciate any feedback( help ) with what I need to do. The problem is a certain amount of money in invested at 8% per year, and 1500$ more than that amount is inve

Question 141103This question is from textbook
: I would really appreciate any feedback( help ) with what I need to do. The problem is a certain amount of money in invested at 8% per year, and 1500$ more than that amount is invested at 9% per year. The annual interest from the 9% investment exceeds the annual interest from the 8% investment by 160$. How much is invested at each rate? This is what I set up.
.08x+.09(x+1500) = 160+??????????? thank you in advance for your imput This question is from textbook

Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you're doing fine __ the 160 is the difference, so just adjust your equation

.09(x+1500)-.08x=160 __ multiply by 100 to clear decimals __ 9x+13500-8x=16000 __ x=2500
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