tracy invested $6000 for one year, part at 10% annual and the balance at 13% annual interest. her total interest for the year was $712.5. how much money did she invested at each rate?

Question

I tried using a table for this I just don't know how to set the problem up.

Anonymous · Answer

Let $x is what she invested at 10% then $(6000-x) is what she invested at 13%. We know that 0.10x + 0.13(6000-x) = 712.5 Solve for x to get the answers.