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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\n" ); document.write( "it has to be one of the following
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\n" ); document.write( "\n" ); document.write( "$2,000
\n" ); document.write( "$3,000
\n" ); document.write( "$4,000
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Algebra.Com's Answer #447211 by mananth(16946)\"\" \"About 
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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
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\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
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\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
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\n" ); document.write( "m.ananth@hotmail.ca
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