# algebra

A total of \$12,000 is invested in two funds paying 9% and 11% simple interest. If the yearly interest is \$1,180, how much of the \$12,000 is invested at each rate?
We have two unknowns: the amount of money invested at 9% and the amount of money invested at 11%.

Sentence (1) ''A total of \$12,000 is invested in two funds paying 9% and 11% simple interest.'' can be restated as (The amount of money invested at 9%) (The amount of money invested at 11%) \$12,000.

Sentence (2) ''If the yearly interest is \$1,180, how much of the \$12,000 is invested at each rate?'' can be restated as (The amount of money invested at 9%) 9% + (The amount of money invested at 11% 11%) total interest of \$1,180.
In shortcut form: (1) x+y=\$12000 (2) .09x+.11y=\$1180

Principal of P dollars = P dollars* Rate in decimal * time in years
Can you show me how you get interest 0.09x on aPrincipal of x dollars?

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x + y = 12000 ---> y = 12000 - x

sub that into the other equation:
.09x + .11y = 1180
.09x+ .11(12000-x) = 1180
.09x+ 1320 - .11x = 1180
-.02x = -140
x = 7000
then y = 12000 - 7000 = 5000

\$7000 was invested at 9%, and \$5000 was invested at 11%

check:
.09(7000) + .11(5000)
= 630 + 550 = 1180

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2. Suppose that \$2000 is invested at a rate of 3.3% ,total amount after 3 years.

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