document.write( "Question 660308: In a two-digit number. the tens digit is 2 more than twice the ones digit. if the digits are reversed and the new number is doubled, the result is 7 less than the original number. What was the original number? \n" ); document.write( "

Algebra.Com's Answer #411117 by kevwill(135) $\"\"$ $\"About$

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Let x by the tens digit in the number, and y be the ones digit. Then we have
\n" ); document.write( "x = 2*y+2 (the tens digit is 2 more than twice the ones digit), and
\n" ); document.write( "2(10*y+x) = (10*x+y) - 7
\n" ); document.write( "Expanding the second equation:
\n" ); document.write( "20*y + 2*x = 10*x + y - 7
\n" ); document.write( "20*y + 2*x - 2*x - y = 10*x + y - 7 - 2*x - y
\n" ); document.write( "19*y = 8*x - 7
\n" ); document.write( "Substituting 2*y+2 for x (from the first equation):
\n" ); document.write( "19*y = 8*(2*y+2) - 7
\n" ); document.write( "19*y = 16*y + 16 - 7
\n" ); document.write( "19*y - 16*y + 9
\n" ); document.write( "3*y = 9
\n" ); document.write( "y = 3
\n" ); document.write( "x = 2*3 + 2 = 8
\n" ); document.write( "So the number we're looking for is 83.
\n" ); document.write( "Reversing the digits give 38; doubling that gives 76, which is 7 less than the original number, 83. \n" ); document.write( "

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