SOLUTION: A part of $6000 was invested at 6% annual interest and the remaining at 7% annual interest. At the end of the year the total received was $6391. How much money was invested at

Question 722697:
A part of $6000 was invested at 6% annual interest and the remaining at 7% annual interest. At the end of the year the total received was $6391. How much money was invested at each rate?
Answer by checkley79(3341) (Show Source): You can put this solution on YOUR website!
P(1+R)^T
X(1+.07)^1+(6000-X)(1+.06)^1=6391
1.07X+(6000-X)*1.06=6391
1.07X+6360-1.06X=6391
.01X=6391-6360
.01X=31
X=31/.01
X=$3,100 INVESTED @ 7%
6000-3100=$2,900 AMOUNT INVESTED @ 6%.
PROOF:
3100*1.07+2900*1.06=6391
3317+3074=6391
6391=6391
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