Question

Maria invested $5000, some at 6% and some at 9%. If she wins $393 annual interest how much is invested in each account.

Answer 1

Ok, so

We could solve this problem using a system of linear equations as follows:

x = amount invested at 6%

y = amount invested at 9%

We are given two numbers in the problem:

$5000 = total money invested in both accounts

$393 = total interest earned in both accounts

Then we could write:

$\begin{cases}x+y=5000 \\ 0.06x+0.09y=393\end{cases}$

This system can be solved using sustitution:

$y=5000-x$

Replacing:

$\begin{gathered} 0.06x+0.09(5000-x)=393 \\ 0.06x+450-0.09x=393 \\ -0.03x=-57 \\ x=1900 \end{gathered}$

And y = 5000 - x, which is 5000 - 1900, and that is 3100.

We obtain that the solution of the system is:

$(x,y)=(1900,3100)$

Therefore, there was invested $3100 in the account at 9% and 1900 in the account at 6%.