document.write( "Question 1094231: The annual interest income on two investments is $79. Of the two investments, $200 more was invested at 5.5% than at 3%. How much was invested at 3% \n" ); document.write( "

Algebra.Com's Answer #708846 by ikleyn(52790) $\"\"$ $\"About$

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document.write( "Let x be the smaller investment, in dollars.\r\n" );
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document.write( "Then the larger investment is x+200 dollars.\r\n" );
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document.write( "Your equation (from the condition) is\r\n" );
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document.write( "0.03*x + 0.055*(x+200) = 79.\r\n" );
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document.write( "Multiply by 1000 both sides to get\r\n" );
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document.write( "30x + 55x + 11000 = 79000  ====>\r\n" );
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document.write( "85x = 79000-11000 = 68000  ====>  x =  = 800.\r\n" );
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document.write( "Answer.  $800 was invested at 3%  and  800+200 = $1000 was invested at 5.5%.\r\n" );
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