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$13,055 is invested, part at 14% and the rest at 9%. If the interest earned from the amount invested at 14% exceeds the interest earned from the amount invested at 9% by $770.62, how much is invested at each rate?

Respuesta :

calculista
let
x-----------> the amount invested at 14%
y-----------> the amount invested at 9%

we know that
x+y=$13,055---------> x=13,055-y-------> equation 1
x*[1.14]=y*[1.09]+770.62------> equation 2

substitute equation 1 in equation 2

1.14*[13,055-y]=1.09y+770.62
14882.7-1.14y=1.09y+770.62
1.14y+1.09y=14882.7-770.62
2.23y=14112.08
y=6,328.29
x=13,055-6,328.29----------> x=6,726.71

the answer is
the amount invested at 14% is $6,726.71
the amount invested at 9% is $6,328.29

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