How much at each rate?

Question

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

Answer 1

$1800 at 9%
$4200 at 11%

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

Substitute, y = 6000 - x

.09x + .11(6000 - x) = 624
.09x + 660 - .11x = 624
-.02x = 36
x = 1800

Answer 2

Let x = the amount invested in the 9% account and then 6,000 - x is the amount invested in the 11% account.

Since i = prt, the xeq is

int earned 9% account + int earned 11% account = 624

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

.09x + 660 - .11x = 624

-.02x = - 36

x = 36/.02 = 1,800 in the 9% account

and 6000-1800 - 4200 in the 11% account

Answer 3

i guess 800-900