document.write( "Question 329117: Pat invested a total of $3000. Part of the money yields 10% interest per year and the rest yields %8 interest per year. If the total yearly interest fromt his investment is $256 how much did Pat invest at 10% and how much did she invest at 8%??? \n" ); document.write( "
Algebra.Com's Answer #235849 by stanbon(75887)\"\" \"About 
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Pat invested a total of $3000. Part of the money yields 10% interest per year and the rest yields %8 interest per year. If the total yearly interest fromt his investment is $256 how much did Pat invest at 10% and how much did she invest at 8%???
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\n" ); document.write( "Equations:
\n" ); document.write( "Quantity Equation: x + y = 3000
\n" ); document.write( "Interest Equation:0.10x + 0.08y = 256
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\n" ); document.write( "Multiply thru the Quantity Eq. by 10
\n" ); document.write( "Multiply thru the Interest Eq. by 100
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\n" ); document.write( "10x + 10y = 30000
\n" ); document.write( "10x + 8y = 25600
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\n" ); document.write( "Subtract and solve for \"y\":
\n" ); document.write( "2y = 4400
\n" ); document.write( "y = $2200 (amt. invested at 8%)
\n" ); document.write( "Since x+y = 3000, x = $800 (amt invested at 10%)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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