document.write( "Question 800981: Peter and Shelly invested $10,000 for one year. Part of it was invested at 10% simple interest and the rest was invested at 6%. At the end of the year the couple had earned exactly $860 in simple interest. How much did they invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #483388 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( " 10000
\n" ); document.write( "Interest----- 860.00
\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= 10000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "10.00% x + 6.00% y= 860
\n" ); document.write( "Multiply by 100
\n" ); document.write( "10 x + 6 y= 86000.00 --------2
\n" ); document.write( "Multiply (1) by -10
\n" ); document.write( "we get
\n" ); document.write( "-10 x -10 y= -100000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -4 y= -14000
\n" ); document.write( "divide by -4
\n" ); document.write( " y = 3500 - x)
\n" ); document.write( "Part I 10.00% $ 6500
\n" ); document.write( "Part II 6.00% $ 3500
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "6500 --------- 10.00% ------- 650.00
\n" ); document.write( "3500 ------------- 6.00% ------- 210.00
\n" ); document.write( "Total -------------------- 860.00
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );