SOLUTION: The total of two amounts of money is $600,000. The larger amount is invested at 8% and the smaller amount at 6%. The total annual income is $44,000. How much is each investment?

Question 909470: The total of two amounts of money is $600,000. The larger amount is invested at 8% and the smaller amount at 6%. The total annual income is $44,000. How much is each investment?
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Total amount of money invested: $600000
x+y=600000,
Total yearly interest for the two accounts is: $44000
0.08*x+0.06*y=44000
x=600000-y
Substitute for x
0.08*(600000-y)+0.06*y=44000
Multiply out
48000-0.08*y+0.06*y=44000
Combine like terms.
-0.02*y=-4000
Isolate y
y=$200000.00 at 6%
x=600000-y
Calculate x
x=$400000.00 at 8%
Check
0.08*400000+0.06*200000=44000
32000+12000=44000
44000=44000
If this statement is TRUE and neither x nor y is negative then all is well
code int1

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