SOLUTION: Bob invested $20,000, part at 14% and part at 13%. If the total interest at the end of the year is $2,720, how much did he invest at 14%?

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Question 55136: Bob invested $20,000, part at 14% and part at 13%. If the total interest at the end of the year is $2,720, how much did he invest at 14%?
Answer by Hook(36) About Me  (Show Source):
You can put this solution on YOUR website!
Lets call x the amount that Bob invested at 14% and y the amount that he invested at 13%.
Combined, Bob invested 20 grand, so we have
x+%2B+y+=+20000
Additionally, we know that the sum of the amount that Bob invested at 14% and 13% is equal to $2720. In math-speak, that says:
.14x%2B.13y+=+2720
So we have the following system of equations:
x+%2B+y+=+20000
.14x+%2B+.13y+=+2720
My strategy will be to use the substitution method.
Starting with
x+%2B+y+=+20000 and solving for x, I get
x+=+20000-y
I'll plop that into the second equation
.14x+%2B+.13y+=+2720
.14%2820000-y%29+%2B+.13y+=+2720
2800-.14y%2B.13y+=+2720
2800+-+.01y+=+2720
-.01y+=+-80
y+=8000
Great! Bob invested $8000 at 13%
I'll put that back into x+=+20000-y to find x
x+=+20000-8000
x+=+12000
So the answer is Bob invested $12000 at 14%
CHECK:
I'll check my work by plugging x = 12000 and y = 8000 into the second equation
.14x%2B.13y+=+2720
.14%2812000%29%2B.13%288000%29+=+2720
1680+%2B+1040+=+2720
2720+=+2720...check!