document.write( "Question 1158613: different amounts are invested at 10%, 11%, and 12% yearly interest. the amount invested at 11% is $300 more than the amount invested at 10% and the total yearly income from all 3 investments is $324. a total of $2900 is invested. find the amount invested at each rate. \n" ); document.write( "

Algebra.Com's Answer #781588 by josgarithmetic(39617) $\"\"$ $\"About$

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document.write( "RATE%      VARIABLE OR QUANT.              EARN\r\n" );
document.write( "10               x                         0.1x\r\n" );
document.write( "11               x+300                     0.11(x+300)\r\n" );
document.write( "12              2900-x-x-300=2600-2x       0.12(2600-2x)\r\n" );
document.write( "TOTALS          2900                       324\r\n" );
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\n" ); document.write( "\n" ); document.write( "Setup and solve the equation $\"0.1x%2B0.11%28x%2B300%29%2B0.12%282600-2x%29=324\"$ .
\n" ); document.write( "Then evaluate the other two quantities. \n" ); document.write( "

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