SOLUTION: two investments are made totaling $10,000. In one year these investments yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and the rest at 6 3/4%. How

Algebra ->  Square-cubic-other-roots -> SOLUTION: two investments are made totaling $10,000. In one year these investments yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and the rest at 6 3/4%. How       Log On


   

Question 443238: two investments are made totaling $10,000. In one year these investments
yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and
the rest at 6 3/4%. How much more money is invested at 6 3/4%?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
two investments are made totaling $10,000. In one year these investments
yield $650 in simple interest. Part of the $10,000 in invested at 5 1/2%, and
the rest at 6 3/4%. How much more money is invested at 6 3/4%?
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2-variable solution:
Quantity Eq. :: x + y = 10,000
Interest Eq. ::0.055x + 0.0675y = 650
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Multiply 1st by 550
Multiply 2nd by 10000
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550x + 550y = 550*10,000
550x + 675y = 6400000
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Subtract and solve for "y":
125y = 9000000
y = $7200 (amt invested at 6 3/4%)
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Cheers,
Stan H.