document.write( "Question 1118391: Part of $6,000 is invested at 12%, another part at 14%, and the remainder at 15% yearly interest. The total yearly income from the three investments is $840. The sum of the amounts invested at 12% and 14% equals the amount invested at 15%. How much is invested at each rate?
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Algebra.Com's Answer #733692 by josgarithmetic(39628) $\"\"$ $\"About$

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document.write( "x    at   12%\r\n" );
document.write( "y    at   14%\r\n" );
document.write( "x+y  at   15%\r\n" );
document.write( "------------------\r\n" );
document.write( "------------------2x+2y=6000  or x+y=3000\r\n" );
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\n" ); document.write( "\n" ); document.write( "Yearly Income, $840
\n" ); document.write( " $\"12x%2B14y%2B15%28x%2By%29=84000\"$ , and $\"x%2By=3000\"$ .\r
\n" ); document.write( "\n" ); document.write( " $\"12x%2B14y%2B15%2A3000=84000\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"12x%2B14y=84000-45000\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"12x%2B14y=39000\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"12x%2B14%283000-x%29=39000\"$ ------from hear, the algebra is a few short simple steps.... \n" ); document.write( "

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