document.write( "Question 672476: An investment advisor invested $18000 in two accounts. One investment earned 11.2% annual simple interest while the other investment lost 4.7%. The total earnings from both investments were $1062. Find the amount invested at 11.2%. \n" ); document.write( "
Algebra.Com's Answer #418056 by mananth(16946)\"\" \"About 
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Part I 11.20% per annum ------------- Amount invested =x
\n" ); document.write( "Part II -4.70% per annum ------------ Amount invested = y
\n" ); document.write( " 18000
\n" ); document.write( "Interest----- 1062
\n" ); document.write( "
\n" ); document.write( "Part I 11.20% per annum ---x
\n" ); document.write( "Part II -4.70% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 18000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "11.20% x + -4.70% y= 1062
\n" ); document.write( "Multiply by 100
\n" ); document.write( "11.2 x + -4.7 y= 106200.00 --------2
\n" ); document.write( "Multiply (1) by -11.2
\n" ); document.write( "we get
\n" ); document.write( "-11.2 x -11.2 y= -201600.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -15.9 y= -95400
\n" ); document.write( "divide by -15.9
\n" ); document.write( " y = 6000
\n" ); document.write( "Part I 11.20% $ 12000
\n" ); document.write( "Part II -4.70% $ 6000
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "12000 --------- 11.20% ------- 1344.00
\n" ); document.write( "6000 ------------- -4.70% ------- -282.00
\n" ); document.write( "Total -------------------- 1062.00
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
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