Question

Larry Mitchel invested part of his $32,000 advance at 7% annual simple interest and the rest at 6% annual simple interest. If this total yearly interest from both accounts was $2,050, find the amount invested at each.

The amount invested at 7% is....
The amount invested at 6% is....

Thank You for the Help

Answer 1

Investment @ 7% = Po
Investment @ 6% = 32,000-Po

Po*r*t + (32000-Po)*r*t = 2,050
Po*0.07*1 + (32000-Po)*0.06*1 = 2050
0.07Po + 1920 - 0.06Po = 2050
0.01Po = 2050-1920 = 130
Po = $13000 @ 7%.

32,000-13000 = $19,000 @ 6%.