SOLUTION: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year. If the total yearly interest from this invest

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year. If the total yearly interest from this invest      Log On


   

Question 269869: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year. If the total yearly interest from this investment is $256, how much did Pat invest at 10 percent and how much at 8 percent?
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
.10x+.08(3,000-x)=256
.10x+240-.08x=256
.02x=256-240
.02x=16
x=16/.02
x=$800 amount invested @ 10%.
3,000-800=$2,200 amount invested @ 8%.
Proof:
.10*800+.08*2,200=256
80+176=256
256=256