document.write( "Question 731665: $6,000 dollars is invested in two different accounts earning 3% and 5% interest. At the end of one year, the two accounts earned $220 in interest. How much money was invested at 3%?
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document.write( "it has to be one of the following
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document.write( "$2,000
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document.write( "$3,000
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document.write( "$4,000 \n" );
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Algebra.Com's Answer #447211 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 3.00% per annum ------------- Amount invested =x \n" ); document.write( "Part II 5.00% per annum ------------ Amount invested = y \n" ); document.write( " 6000 \n" ); document.write( "Interest----- 220.00 \n" ); document.write( " \n" ); document.write( "Part I 3.00% per annum ---x \n" ); document.write( "Part II 5.00% per annum ---y \n" ); document.write( "Total investment \n" ); document.write( "x + 1 y= 6000 -------------1 \n" ); document.write( "Interest on both investments \n" ); document.write( "3.00% x + 5.00% y= 220 \n" ); document.write( "Multiply by 100 \n" ); document.write( "3 x + 5 y= 22000.00 --------2 \n" ); document.write( "Multiply (1) by -3 \n" ); document.write( "we get \n" ); document.write( "-3 x -3 y= -18000.00 \n" ); document.write( "Add this to (2) \n" ); document.write( "0 x 2 y= 4000 \n" ); document.write( "divide by 2 \n" ); document.write( " y = 2000 \n" ); document.write( "Part I 3.00% $ 4000 \n" ); document.write( "Part II 5.00% $ 2000 \n" ); document.write( " \n" ); document.write( "CHECK \n" ); document.write( "4000 --------- 3.00% ------- 120.00 \n" ); document.write( "2000 ------------- 5.00% ------- 100.00 \n" ); document.write( "Total -------------------- 220.00 \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |