A total of $8000 is invested: part at 7% and the remainder at

11%. How much is invested at each rate if the annual interest is $760?

1.).07x +.11y = 760

2.) x + y=8000

we can solve the second equation for either x or y

y = 8000 - x

or

x=8000 - y

plug 1 of those in to equation 1 (I'll use y=8000 - x)

.07x +.11 (8000 - x) = 760

.07x + 880 -.11x = 760

-.04x + 880 = 760

-.04x = - 120

x=3000

So 3000 is invested in 7%

5000 is invested in 11%

Check

(3000 x. 07) + (5000 x. 11) = 720

210 + 550 = 760

Success!