SOLUTION: Hi. I'm having a lot of trouble with this question. Suppose you invested $5000 in three different funds for one year. The funds paid simple interest of 8%, 10%, and 7%, respectivel

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Question 174521This question is from textbook Algebra 2
: Hi. I'm having a lot of trouble with this question. Suppose you invested $5000 in three different funds for one year. The funds paid simple interest of 8%, 10%, and 7%, respectively. The total interest at the end of one year was $405. You invested $500 more at 10% than at 8%. How much did you invest in the 10% fund?
The answer of the book is $1500, but I really need to know how to get there and I would really appreciate it your help. THANKS!
This question is from textbook Algebra 2

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the amount invested at 10%, then x-500 = the amount invested at 8%, so that 2x-500 would be the amount invested at 7%.
The sum of the three invested amounts (x)+(x-500)+(2x-500) is $5,000.
Now let's look at the amounts of interest earned on these amounts:
x%280.1%29 is the amount earned at 10 %.
%28x-500%29%280.08%29 is the amount earned at 8%.
2x-500%29%280.07%29 is the amount earned at 7%.
The sum of these three amounts is given as $405.
So we can write the following equation in x.
0.1x%2B0.08%28x-500%29%2B+0.07%282x-500%29+=+405 Simplify and solve for x.
0.1x%2B0.08x-40%2B0.14x-35+=+405
0.32x-75+=+405 Add 75 to both sides.
0.32x+=+480 Divide both sides by 0.32
highlight%28x+=+1500%29
$1,500 was invested at 10%
Check:
x%2B%28x-500%29%2B%282x-500%29+=+5000
1500%2B%281500-500%29%2B%283000-500%29+=+5000
1500%2B1000%2B2500+=+5000
5000+=+5000