Larry Mitchell invested part of his 13000 advance at 3 annua

Larry Mitchell invested part of his $13,000 advance at 3% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $1,110, find the amount invested at each rate.

Solution

Let amount investes at 3% be ---x

amount invested at 9% be ---- 13000 -x

One year interest -----$ 1110

Simple interest = P*R*T/100

1110 = x*3/100 + ( 13000-x)9/100

1110 = 0.03x -0.09x + 1170

-60 = -0.06x

x =$ 1000 invested at 3% rate

13000-1000 = $12000 at 9% rate