SOLUTION: A man has $800 more invested at 4% than at 3%. If the total interest for the two investments is $130. How much is invested at each rate?

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Question 1038878: A man has $800 more invested at 4% than at 3%. If the total interest for the two investments is $130. How much is invested at each rate?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

x = amount invested at 4%
y = amount invested at 3%


"A man has $800 more invested at 4% than at 3%" so we know that x=y%2B800

The amount of interest earned, in one year, for the 4% account is x%2A0.04+=+0.04x
The amount of interest earned, in one year, for the 3% account is y%2A0.03+=+0.03y

Put together, the total interest is 0.04x%2B0.03y which is set equal to 130 because this is the total interest for the year.

We have 0.04x%2B0.03y=130

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Use the two equations 0.04x%2B0.03y=130 and x=y%2B800 to solve for x and y. Start by using substitution.


0.04x%2B0.03y=130

0.04%28y%2B800%29%2B0.03y=130 Plug in x=y%2B800. Solve for y

0.04y%2B0.04%2A%28800%29%2B0.03y=130

0.04y%2B32%2B0.03y=130

0.07y%2B32=130

0.07y%2B32-32=130-32

0.07y=98

0.07y%2F0.07=98%2F0.07

y=1400

Now that we know that y=1400, use this to find x.

x=y%2B800

x=1400%2B800

x=2200


So we know that x=2200 and y=1400


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Final Answer:

$2,200 was invested at 4%
$1,400 was invested at 3%