document.write( "Question 535038: $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 5%? \r
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Algebra.Com's Answer #351739 by mananth(16946) You can put this solution on YOUR website! Investment I 3.00% per annum ---x \n" ); document.write( "Investment II 5.00% per annum ---y \n" ); document.write( " \n" ); document.write( "x + y= 6000 ------------------------1 \n" ); document.write( "3.00% x + 5.00% y= = $220.00 \n" ); document.write( "Multiply by 100 \n" ); document.write( "3 x + 5 y= = $22,000.00 --------2 \n" ); document.write( " \n" ); document.write( "Multiply (1) by -3 \n" ); document.write( "we get \n" ); document.write( " \n" ); document.write( "-3 x -3 y= = -18000.00 \n" ); document.write( " \n" ); document.write( "Add this to (2) \n" ); document.write( " \n" ); document.write( "0 x+ 2 y= = $4,000.00 \n" ); document.write( " \n" ); document.write( "divide by 2 \n" ); document.write( " \n" ); document.write( "y = $2,000.00 \n" ); document.write( "investment at 5.00% \n" ); document.write( "Balance $4,000.00 investment at 3.00% \n" ); document.write( "CHECK \n" ); document.write( "$4,000.00 --------- 3.00% ------- $120.00 \n" ); document.write( "$2,000.00 ------- 5.00% ------- $100.00 \n" ); document.write( "Total -------------- $220.00 \r \n" ); document.write( "\n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |