document.write( "Question 319312: A person invests $8000 for one year. Part of this amount is invested at 2.65% simple interest and the rest is invested at 3.12% simple interest. If the total interest received from both investments is $235.50, find the amounts invested at each rate \n" ); document.write( "
Algebra.Com's Answer #228666 by stanbon(75887)\"\" \"About 
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A person invests $8000 for one year. Part of this amount is invested at 2.65% simple interest and the rest is invested at 3.12% simple interest. If the total interest received from both investments is $235.50, find the amounts invested at each rate.
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\n" ); document.write( "Equations:
\n" ); document.write( "Quantity Equation: x + y = 8000
\n" ); document.write( "Interest Equation:0.0265x + 0.0312y = 235.50
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\n" ); document.write( "Multiply thru the 1st equation by 265
\n" ); document.write( "Multiply thru the 2nd equation by 10000
\n" ); document.write( "265x + 265y = 265*8000
\n" ); document.write( "265x + 312y = 2355000
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\n" ); document.write( "Subtract 1st Eq. from 2nd and solve for \"y\":
\n" ); document.write( "47y = 235000
\n" ); document.write( "y = $5000 (amt. invested at 2.65%)
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\n" ); document.write( "x = 8000-5000 = $3000 (amt. invested at 3.12%)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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