document.write( "Question 1000227: A total of $5000 is invested at 3% and 5% annual interest. After 1 year, the total interest equals $210. How much money is invested at each interest rate? \n" ); document.write( "

Algebra.Com's Answer #617716 by mananth(16946) $\"\"$ $\"About$

You can put this solution on YOUR website!
Part I 3.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 5.00% per annum ------------ Amount invested = y = 3 x
\n" ); document.write( " 5000
\n" ); document.write( "Interest----- 210
\n" ); document.write( "
\n" ); document.write( "Part I 3.00% per annum ---x *
\n" ); document.write( "Part II 5.00% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 5000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "3.00% x + 5.00% y= 210
\n" ); document.write( "Multiply by 100
\n" ); document.write( "3 x + 5 y= 21000.00 --------2
\n" ); document.write( "Multiply (1) by -3
\n" ); document.write( "we get
\n" ); document.write( "-3 x -3 y= -15000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x 2 y= 6000
\n" ); document.write( "divide by 2
\n" ); document.write( " y = 3000 -
\n" ); document.write( "Part I 3.00% $ 2000
\n" ); document.write( "Part II 5.00% $ 3000
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "2000 --------- 3.00% ------- 60.00
\n" ); document.write( "3000 ------------- 5.00% ------- 150.00
\n" ); document.write( "Total -------------------- 210.00
\n" ); document.write( "
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "

\n" );