document.write( "Question 1166223: The sum of the digits of a certain two-digit number is 10. If the digits are reversed, a new number is formed which is one less than twice the original number. \n" ); document.write( "

Algebra.Com's Answer #790706 by math_helper(2461) $\"\"$ $\"About$

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document.write( "The original number:  10a + b   (a and b are each single digits from the set {'0','1',...,'9'}\r
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document.write( "Reversing the digits gives the number  10b + a\r
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document.write( "Swapped number = 2(Original number) - 1:
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document.write( "10b + a = 2(10a + b) - 1 \r
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document.write( "Because a+b=10, we can replace 'a' with '10-b' to get one equation in one unknown (in this case, b):
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document.write( "10b + (10-b) = 2(10(10-b) + b) - 1
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document.write( "...
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document.write( "This simplifies to b=7 ==> a=3\r
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document.write( "The original number is 37
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document.write( "The swapped number is  73\r
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document.write( "Check:
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document.write( "73 = 2(37)-1 = 74-1  (ok)
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