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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?

Sagot :

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!
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