SOLUTION: Larry Mitchell invested part of his $40,000 advance at 5% annual simple interest and the rest at 7% annual simple interest. If his total yearly interest from both accounts was $2,2

Question 908700: Larry Mitchell invested part of his $40,000 advance at 5% annual simple interest and the rest at 7% annual simple interest. If his total yearly interest from both accounts was $2,280 find the amount invested at each rate.
The amount invested at 5% is?
The amount invested at 7% is?
Found 2 solutions by ewatrrr, richwmiller:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
x amount at 7%
.07x + .05(40000-x) = 2280
Solve for x
x = 280/.02
and then find
$40,000 - x for the 5% amount
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Don't waste your time nor our time writing this nonsense.
The amount invested at 5% is?
The amount invested at 7% is?
It serves no useful purpose and it is annoying. Rather,spend time working on the problem.
Total amount of money invested: $40000
x+y=40000,
Total yearly interest for the two accounts is: $2280
0.05*x+0.07*y=2280
x=40000-y
Substitute for x
0.05*(40000-y)+0.07*y=2280
Multiply out
2000-0.05*y+0.07*y=2280
Combine like terms.
0.02*y=280
Isolate y
y=$14000.00 at 7%
x=40000-y
Calculate x
x=$26000.00 at 5%
Check
0.05*26000+0.07*14000=2280
1300+980=2280
2280=2280
If this statement is TRUE and neither x nor y is negative then all is well

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