SOLUTION: You invested $10,000 in two accounts, one paying 8% and the other account paying 10% annual interest. At the end of the year, the total interest from these investments was $940. Ho

Question 970109: You invested $10,000 in two accounts, one paying 8% and the other account paying 10% annual interest. At the end of the year, the total interest from these investments was $940. How much was invested at each rate?
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
Amount in Account 1 is x
Amount in Account 2 is $10,000-x; Their sum is $10,000
Account 1 pays .08x per year I=PRT
Account 2 pays (.1)(10.000-x) per year
Their sum is $940
.08x + 1000-.10x=940 We distribute the .1 over the (10,000-x)
Collect terms and subtract 1000 from both sides
-0.02x=-$60
Divide by (-1)
0.02 x=60
Multiply by 50 (or divide by 0.02)
x=3000 @ 8%, $240/year ACCOUNT 1
10,000-x =$7000 @ 10%, $740 year ACCOUNT 2
Their sum is $940.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
You invested $10,000 in two accounts, one paying 8% and the other account paying 10% annual interest. At the end of the year, the total interest from these investments was $940. How much was invested at each rate?
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Equation:
interest + interest = interest
0.08*x + 0.10(10000-x) = 940
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8x + 10*10000 - 10x = 94000
-2x = -6000
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x = $3000 (amt. invested at 8%)
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10,000-x = $7,000 (amt. invested at 10%)
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Cheers,
Stan H.
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