View Full Version : What equation do I use to solve this?
tywatson
Jan 19, 2008, 10:35 AM
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?
PolluxCastor
Jan 19, 2008, 10:56 AM
He should have invested it all at 11%, he would have made $660.
galactus
Jan 19, 2008, 02:01 PM
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.
Ronniesha
Jan 23, 2008, 10:48 PM
4y-14=10
tywatson
Jan 24, 2008, 01:31 PM
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
PolluxCastor
Jan 24, 2008, 05:21 PM
Yay!