Question 141500
Let x be the tens digit.
Let y be the ones digit.


The number is then {{{10x + y}}} 
(if x were 6 and y were 3, then the number would be 63 which is {{{10(6)+3}}})


and we know that {{{x = 2y}}}


The new number with the digits reversed must be {{{10y+x}}} and this is 36 less than the original number, so:


{{{10x+y-36=10y+x}}}.  To solve, substitute {{{2y}}} for {{{x}}} because {{{x=2y}}}, and then solve for {{{y}}}, multiply by 2 to get {{{x}}}, and then construct the number.


Remember to check your work by constructing the number, reversing the digits, and then subtracting to make sure you get 36.