# Tracy invested 6000 dollars for 1 year, part at 10% annual interest and the balance at 13% annual interest. Her total interest for the year is 712.50 dollars. How much money did she invest at each rate?

Nov 7, 2015

$2250 @10%$3750 @13%

#### Explanation:

Let $x$ be the amount invested at 10%
$\implies 6000 - x$ is the amount invested at 13%

$0.10 x + 0.13 \left(6000 - x\right) = 712.50$

$\implies 10 x + 13 \left(6000 - x\right) = 71250$

$\implies 10 x + 78000 - 13 x = 71250$
$\implies - 3 x + 78000 = 71250$

$\implies 3 x = 78000 - 71250$

$\implies 3 x = 6750$
$\implies 2250$

$\implies 6000 - x = 3750$