Question

A self-employed contractor nearing retirement made two investments totaling 15,000. In one year, these investments yielded $1023 in simple interest. Part of the money was invested at 6% and the rest at 7.5%. How much was invested at each rate? show work

Answer 1

Let x=investment at 6%
Let y=investment at 7.5%

x+y=15,000
0.06x+0.075y=1023 Multiply the first equation by 0.075 to eliminate y and find x

0.075x+0.075y=1125
0.06x+0.075y=1023 Subtract this equation from the first one to get:

0.015x=102
x=6800 Substitute this value into one of the equations--I used:

x+y=15,000
y=15,000-6800=8200

Therefore, $6800 is invested at 6% and $8200 is invested at 7.5%