document.write( "Question 862095: the sum of the digits of a two digit number is 9, when the digits are reversed the new number is 27 less than the original number \n" ); document.write( "

Algebra.Com's Answer #519463 by mananth(16946) $\"\"$ $\"About$

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x= number in ten's place
\n" ); document.write( "y= number in units place
\n" ); document.write( "
\n" ); document.write( "1 x + 1 y = 9 .............1 \r
\n" ); document.write( "\n" ); document.write( "10y+x = 10x+y+27
\n" ); document.write( "9y-9x=27\r
\n" ); document.write( "\n" ); document.write( "x= number in ten's place
\n" ); document.write( "y= number in units place
\n" ); document.write( "
\n" ); document.write( "1 x + 1 y = 9 .............1
\n" ); document.write( "Total value
\n" ); document.write( "-9 x + 9 y = 27 .............2
\n" ); document.write( "Eliminate y
\n" ); document.write( "multiply (1)by -9
\n" ); document.write( "Multiply (2) by 1
\n" ); document.write( "-9 x -9 y = -81
\n" ); document.write( "-9 x + 9 y = 27
\n" ); document.write( "Add the two equations
\n" ); document.write( "-18 x = -54
\n" ); document.write( "/ -18
\n" ); document.write( "x = 3
\n" ); document.write( "plug value of x in (1)
\n" ); document.write( "1 x + 1 y = 9
\n" ); document.write( "3 + y = 9
\n" ); document.write( " y = 9 -3
\n" ); document.write( " y = 6
\n" ); document.write( " y = 6
\n" ); document.write( "x= 3.00 number in ten's place
\n" ); document.write( "y= 6.00 number in units place \r
\n" ); document.write( "\n" ); document.write( "The number is 36
\n" ); document.write( "m.ananth@hotmail.ca
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