SOLUTION: A man invests $350,000, part of it at 9% and the other part at 12%. How much did he invest at each rate if he receives in one year a total interest of $36,000?
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: A man invests $350,000, part of it at 9% and the other part at 12%. How much did he invest at each rate if he receives in one year a total interest of $36,000?
Log On
Question 909467: A man invests $350,000, part of it at 9% and the other part at 12%. How much did he invest at each rate if he receives in one year a total interest of $36,000? Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! Total amount of money invested: $350000
x+y=350000,
Total yearly interest for the two accounts is: $36000
0.09*x+0.12*y=36000
x=350000-y
Substitute for x
0.09*(350000-y)+0.12*y=36000
Multiply out
31500-0.09*y+0.12*y=36000
Combine like terms.
0.03*y=4500
Isolate y
y=$150000.00 at 12%
x=350000-y
Calculate x
x=$200000.00 at 9%
Check
0.09*200000+0.12*150000=36000
18000+18000=36000
36000=36000
If this statement is TRUE and neither x nor y is negative then all is well
code int1
