document.write( "Question 376455: The sum of three digits is 5. The first and last digits are the same. If the middle digit is exchanged with the first digit, the new number is 90 less than the original number. Find the original number. \n" ); document.write( "
Algebra.Com's Answer #267832 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
The original number: 100x + 10y + z\r
\n" ); document.write( "\n" ); document.write( "The sum of three digits is 5.
\n" ); document.write( "x + y + z = 5
\n" ); document.write( ":
\n" ); document.write( " The first and last digits are the same.
\n" ); document.write( "x = z
\n" ); document.write( "therefore we can write the number
\n" ); document.write( "100x + 10y + x
\n" ); document.write( ":
\n" ); document.write( " If the middle digit is exchanged with the first digit, the new number is 90 less than the original number.
\n" ); document.write( ":
\n" ); document.write( "100x + 10y + x = 100y + 10x + x + 90
\n" ); document.write( "combine like terms
\n" ); document.write( "100x - 10x + x - x = 100y - 10y + 90
\n" ); document.write( "90x = 90y + 90
\n" ); document.write( "simplify, divide by 90
\n" ); document.write( "x = y + 1
\n" ); document.write( ":
\n" ); document.write( "x + y + z = 5
\n" ); document.write( "Since 1st and last digit are the same
\n" ); document.write( "(y+1) + y + (y+1) = 5
\n" ); document.write( "3y = 5 - 2
\n" ); document.write( "y = 1
\n" ); document.write( "then, x and z = 2
\n" ); document.write( ":
\n" ); document.write( "212 is the original number
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\n" ); document.write( "Check in the statement;
\n" ); document.write( "If the middle digit is exchanged with the first digit, the new number is 90 less than the original number.
\n" ); document.write( "212 = 122 + 90; confirms our solution\r
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