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 $100 more than the interest earned by the 11% investment, find the amount invested at each rate. \r
\n" );
document.write( "
\n" );
document.write( "\n" );
document.write( "Can you show work?? \n" );
document.write( "
Algebra.Com's Answer #336748 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 11.00% per annum ------------- x \n" ); document.write( "Part II 14.00% per annum ------------ y \n" ); document.write( " total investment 2450 \n" ); document.write( " Interest----- 11.00% x \n" ); document.write( " Interest = 14.00% y greater by 100 \n" ); document.write( " 11.00% x = 14.00% y + 100 \n" ); document.write( " multiply by 100 \n" ); document.write( " 11 x - 14 y = 10000 --------1 \n" ); document.write( "x + y = 2450 -------- 2 \n" ); document.write( "eliminate y \n" ); document.write( "Multiply 1 by 1 \n" ); document.write( "Multiply 2 by 14 \n" ); document.write( "we get \n" ); document.write( "11 x -14 y = 10000 \n" ); document.write( "14 x +14 y = 34300 \n" ); document.write( "25 x = 44300 \n" ); document.write( "/ 25 \n" ); document.write( " x= 1772 \n" ); document.write( "Plug the value of x in in 1 \n" ); document.write( "19492 -14 y = 10000 \n" ); document.write( " -14 y = 10000 -19492 \n" ); document.write( " -14 y = -9492 \n" ); document.write( " / -14 \n" ); document.write( " y= 678 \n" ); document.write( "$ 1772 at 11.00% \n" ); document.write( "$ 678 at 14.00% \n" ); document.write( " \n" ); document.write( " |