document.write( "Question 547066: how can 56000 be invested part at 4% annual simple interest ant the remainder at 10% annual simple interest so that the interest erned by the two accounts is equal at the end if the year \n" ); document.write( "

Algebra.Com's Answer #356194 by mananth(16946) $\"\"$ $\"About$

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x+y=56000
\n" ); document.write( "x@ 4%
\n" ); document.write( "y@10%
\n" ); document.write( "0.04x = 0.1y
\n" ); document.write( "0.04x-0.1y=0\r
\n" ); document.write( "\n" ); document.write( "x+y=56000 .............1
\n" ); document.write( "0.04x-0.1y=0 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by 1
\n" ); document.write( "Multiply (2) by 10
\n" ); document.write( "1 x 1 y = 56000
\n" ); document.write( "0.4 x + -1 y = 0
\n" ); document.write( "Add the two equations
\n" ); document.write( "1.4 x = 56000.000
\n" ); document.write( "/ 1.4
\n" ); document.write( "x = 40000
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1 x + 1 y = 56000
\n" ); document.write( "40000 + 1 y = 56000
\n" ); document.write( " 1 y = 56000 -40000
\n" ); document.write( " 1 y = 16000
\n" ); document.write( " y = 16000 \r
\n" ); document.write( "\n" ); document.write( "4000@4%
\n" ); document.write( "16000@10%
\n" ); document.write( " \n" ); document.write( "

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