document.write( "Question 357470: When the digits of a two-digit number are reversed, the new number is 9 more than the original number, and the sum of the digits of the original number is 13. What is the original number? \n" ); document.write( "
Algebra.Com's Answer #255277 by ankor@dixie-net.com(22740)\"\" \"About 
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When the digits of a two-digit number are reversed, the new number is 9 more
\n" ); document.write( "than the original number, and the sum of the digits of the original number is 13.
\n" ); document.write( "What is the original number?
\n" ); document.write( ":
\n" ); document.write( "x = the 10s digit, y = the units
\n" ); document.write( "then
\n" ); document.write( "10x + y = the original number
\n" ); document.write( "and
\n" ); document.write( "10y + x = the reversed number
\n" ); document.write( ":
\n" ); document.write( "Rev number = orig number + 9
\n" ); document.write( "10y + x = 10x + y + 9
\n" ); document.write( "10y - y = 10x - x + 9
\n" ); document.write( "9y = 9x + 9
\n" ); document.write( "simplify, divide by 9
\n" ); document.write( "y = x + 1
\n" ); document.write( ":
\n" ); document.write( "\"the sum of the digits of the original number is 13. \"
\n" ); document.write( "(kind of silly, the sum of the digits is the same original or reversed, right?\"
\n" ); document.write( "anyway
\n" ); document.write( "x + y = 13
\n" ); document.write( ":
\n" ); document.write( "Rearrange the y = x + 1 and add to the above equation
\n" ); document.write( "-x + y = 1
\n" ); document.write( "+x + y = 13
\n" ); document.write( "------------addition eliminates x, find y
\n" ); document.write( "2y = 14
\n" ); document.write( "y = 7
\n" ); document.write( ":
\n" ); document.write( "I'll let you find x, and check it in the 1st equation \n" ); document.write( "
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