1. ## Interest Calculation

You invested \$11,000 in 2 accounts paying 4% and 8% annual interest. If the total interest earned for the year was \$760, how much was invested at (a) 4% and (b) 8%?

Thanks for the help!

2. Originally Posted by bearhug

You invested \$11,000 in 2 accounts paying 4% and 8% annual interest. If the total interest earned for the year was \$760, how much was invested at (a) 4% and (b) 8%?

Thanks for the help!
Let $n$ be the amount invested in the account with 4%. Therefore $11000-n$ is invested in the account with 8%

$I_{4\%}(1) = (0.04n)^1 = 0.04n$

$I_{8\%}(1) = [(11000-n)(0.08)]^1 = 880-0.08n$

These totals are equal to \$760. So we add them to give:

$880-0.08n+0.04n = 760$

Solve for n which will be the answer to part a. To find the answer to part b subtract the value of n from 11,000

$-0.04n = -120$

$n = \\,3000$

Therefore \$3,000 is invested at 4% and \$8,000 is invested at 8%

I am not 100% sure about the position of n regarding the power. In this case it's not important for the power 1 doesn't change anything