document.write( "Question 750078: A man has invested $8000, part of 10% and the remainder at 6% simple interest. how much has he invested at each rate if his annual income from these investments is $536 \n" ); document.write( "
Algebra.Com's Answer #456340 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
Part I 10.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 6.00% per annum ------------ Amount invested = y
\n" ); document.write( " 8000
\n" ); document.write( "Interest----- 536.00
\n" ); document.write( "
\n" ); document.write( "Part I 10.00% per annum ---x
\n" ); document.write( "Part II 6.00% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 8000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "10.00% x + 6.00% y= 536
\n" ); document.write( "Multiply by 100
\n" ); document.write( "10 x + 6 y= 53600.00 --------2
\n" ); document.write( "Multiply (1) by -10
\n" ); document.write( "we get
\n" ); document.write( "-10 x -10 y= -80000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -4 y= -26400
\n" ); document.write( "divide by -4
\n" ); document.write( " y = 6600
\n" ); document.write( "Part I 10.00% $ 1400
\n" ); document.write( "Part II 6.00% $ 6600
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "1400 --------- 10.00% ------- 140.00
\n" ); document.write( "6600 ------------- 6.00% ------- 396.00
\n" ); document.write( "Total -------------------- 536.00
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );