document.write( "Question 497529: A sum of $2,450 is invested, part of it at 11% interest and the remainder at 14%. If the interest earned by the 14% investment is $100 more than the interest earned by the 11% investment, find the amount invested at each rate. \r
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\n" ); document.write( "\n" ); document.write( "Can you show work??
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Algebra.Com's Answer #336748 by mananth(16946)\"\" \"About 
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Part I 11.00% per annum ------------- x
\n" ); document.write( "Part II 14.00% per annum ------------ y
\n" ); document.write( " total investment 2450
\n" ); document.write( " Interest----- 11.00% x
\n" ); document.write( " Interest = 14.00% y greater by 100
\n" ); document.write( " 11.00% x = 14.00% y + 100
\n" ); document.write( " multiply by 100
\n" ); document.write( " 11 x - 14 y = 10000 --------1
\n" ); document.write( "x + y = 2450 -------- 2
\n" ); document.write( "eliminate y
\n" ); document.write( "Multiply 1 by 1
\n" ); document.write( "Multiply 2 by 14
\n" ); document.write( "we get
\n" ); document.write( "11 x -14 y = 10000
\n" ); document.write( "14 x +14 y = 34300
\n" ); document.write( "25 x = 44300
\n" ); document.write( "/ 25
\n" ); document.write( " x= 1772
\n" ); document.write( "Plug the value of x in in 1
\n" ); document.write( "19492 -14 y = 10000
\n" ); document.write( " -14 y = 10000 -19492
\n" ); document.write( " -14 y = -9492
\n" ); document.write( " / -14
\n" ); document.write( " y= 678
\n" ); document.write( "$ 1772 at 11.00%
\n" ); document.write( "$ 678 at 14.00%
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