SOLUTION: I have been at this math problem for quiet a while now. I would really like it if you can help me out. The answer may be so easy to get, but I just can't figure it out. Please help

Question 179512: I have been at this math problem for quiet a while now. I would really like it if you can help me out. The answer may be so easy to get, but I just can't figure it out. Please help.
The word problem is as followed:
Mr. X invested an amount for 7% and twice that amount for 8%. At the end of the year the income was $11500. What was the amount invested at 7%? amount invested at 8%?

Thanks so much .. your help is greatly appreciated!

Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(48551) (Show Source):
You can put this solution on YOUR website!
Mr. X invested an amount for 7% and twice that amount for 8%. At the end of the year the income was $11500. What was the amount invested at 7%? amount invested at 8%?
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Equation:
interest + interest = 11500
0.07x + 0.08(2x) = 11500
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Multiply thru by 100 to get:
7x + 16x = 1,150,000
23x = 1,150,000
x = $50,000 (amt. invested at 7%)
2x = $100,000 (amt. invested at 8%)
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Cheers,
Stan H.

Answer by ankor@dixie-net.com(12699) (Show Source):
You can put this solution on YOUR website!
Mr. X invested an amount for 7% and twice that amount for 8%. At the end of the
year the income was $11500. What was the amount invested at 7%? amount invested
at 8%?
:
let x = amt invested at 7%
then
2x = amt invested at 8%
;
.07x + .08(2x) = 11500
.07x + .16x = 11500
.23x = 11500
x =
x = $50,000 invested at 7%
and, obviously, $100,000 invested at 8%
;
;
Check by finding the interest of each
.07 * 50000 = 3500
.08 * 100000= 8000
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for a total = 11500