document.write( "Question 1088591: John invests $6000, part at 12% interest and part at 8%. If he earned a total of $660 in interest for the year, how much did he invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #702889 by Tatiana_Stebko(1539) $\"\"$ $\"About$

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Let $\"x\"$ be the amount of money invested at 12%
\n" ); document.write( "Let $\"y\"$ be the amount of money invested at 8%
\n" ); document.write( "Since John invests $6000, then $\"x%2By=6000\"$
\n" ); document.write( "He earned a total of $660 in interest, hence $\"0.12x%2B0.08y=660\"$
\n" ); document.write( "The system of equations is
\n" ); document.write( " $\"system%28x%2By=6000%2C+0.12x%2B0.08y=660%29\"$
\n" ); document.write( " $\"system%28y=6000-x%2C+0.12x%2B0.08%286000-x%29=660%29\"$
\n" ); document.write( " $\"system%28y=6000-x%2C+0.12x%2B480-0.08x=660%29\"$
\n" ); document.write( " $\"system%28y=6000-x%2C+0.04x=180%29\"$
\n" ); document.write( " $\"system%28y=6000-x%2C+x=180%2F0.04%29\"$
\n" ); document.write( " $\"system%28y=6000-4500%2C+x=4500%29\"$
\n" ); document.write( " $\"system%28y=1500%2C+x=4500%29\"$
\n" ); document.write( "John invested $4500 at 12% and $1500 at 8%
\n" ); document.write( "Answer: $4500 at 12% and $1500 at 8% \n" ); document.write( "

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