document.write( "Question 732769: $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%? \n" ); document.write( "
Algebra.Com's Answer #448034 by mananth(16946)\"\" \"About 
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\n" ); document.write( "Part I 5.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 3.00% per annum ------------ Amount invested = y
\n" ); document.write( " 6000
\n" ); document.write( "Interest----- 220.00
\n" ); document.write( "
\n" ); document.write( "Part I 5.00% per annum ---x
\n" ); document.write( "Part II 3.00% 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( "5.00% x + 3.00% y= 220
\n" ); document.write( "Multiply by 100
\n" ); document.write( "5 x + 3 y= 22000.00 --------2
\n" ); document.write( "Multiply (1) by -5
\n" ); document.write( "we get
\n" ); document.write( "-5 x -5 y= -30000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -2 y= -8000
\n" ); document.write( "divide by -2
\n" ); document.write( " y = 4000
\n" ); document.write( "Part I 5.00% $ 2000
\n" ); document.write( "Part II 3.00% $ 4000
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "2000 --------- 5.00% ------- 100.00
\n" ); document.write( "4000 ------------- 3.00% ------- 120.00
\n" ); document.write( "Total -------------------- 220.00
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\n" ); document.write( "m.ananth@hotmail.ca
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