Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate?

He should have invested it all at 11%, he would have made $660.

Let x = the amount invested at 9%, then we have .09x

Then the amount invested at 11% would be 0.11(6000-x)

This must equal 624.

So, we have the equation:

.09x+.11(6000-x)=624

Solve for x.

4y-14=10

x + y = 6000
.09x + .11y = 624

Multiply by .09 and subtract

.09x + .11y = 624
- .09x - .09y = 540
__________________
.02y = 84

y = 84 / .02

y = 4200

x + 4200 = 6000
x = 6000-4200
x = 1800

Amount invested at 9% = 1800
Amount invested at 11% = 4200

Yay!