SOLUTION: matt invested $3200, part at 6% and part at 12%. after one year, he earned $316 worth of interest. how much did he invest at 12% ? (round to the nearest cent)
Algebra ->
Customizable Word Problem Solvers
-> Finance
-> SOLUTION: matt invested $3200, part at 6% and part at 12%. after one year, he earned $316 worth of interest. how much did he invest at 12% ? (round to the nearest cent)
Log On
Question 911739: matt invested $3200, part at 6% and part at 12%. after one year, he earned $316 worth of interest. how much did he invest at 12% ? (round to the nearest cent) Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! Total amount of money invested: $3200
x+y=3200,
Total yearly interest for the two accounts is: $316
0.06*x+0.12*y=316
x=3200-y
Substitute for x
0.06*(3200-y)+0.12*y=316
Multiply out
192-0.06*y+0.12*y=316
Combine like terms.
0.06*y=124
Isolate y
y=$ 2066.67 at 12% earns $248.0 interest
x=3200-y
Calculate x
x=$ 1133.33 at 6% earns $68.0 interest
Check
0.06*1133.33333+0.12*2066.66667=316
68.0+248.0=316
316.0=316
If 316.0=316 is TRUE and neither x nor y is negative then all is well
codeint1