document.write( "Question 899731: Joe received a bonus of $20000 at the end of the year for exceptional performance on the job. He decides to invest the money into two different investment vehickles, one paying 5% interest, the other 6.5% interest. At the end of the year, the total interest earned from both accounts, together, was $1240. Find the amount invested in each account. \n" ); document.write( "
Algebra.Com's Answer #545575 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
Part I 5.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 6.50% per annum ------------ Amount invested = y
\n" ); document.write( " 20000
\n" ); document.write( "Interest----- 1240.00
\n" ); document.write( "
\n" ); document.write( "Part I 5.00% per annum ---x
\n" ); document.write( "Part II 6.50% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 20000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "5.00% x + 6.50% y= 1240
\n" ); document.write( "Multiply by 100
\n" ); document.write( "5 x + 6.5 y= 124000.00 --------2
\n" ); document.write( "Multiply (1) by -5
\n" ); document.write( "we get
\n" ); document.write( "-5 x -5 y= -100000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x 1.5 y= 24000
\n" ); document.write( "divide by 1.5
\n" ); document.write( " y = 16000
\n" ); document.write( "Part I 5.00% $ 4000
\n" ); document.write( "Part II 6.50% $ 16000
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "4000 --------- 5.00% ------- 200.00
\n" ); document.write( "16000 ------------- 6.50% ------- 1040.00
\n" ); document.write( "Total -------------------- 1240.00
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( "
\n" ); document.write( " \n" ); document.write( "
\n" );