Question 515818
Let {{{a}}} = the tens digit
Let {{{b}}} = the ones digit
given:
{{{ a = 2b - 2 }}}
{{{ a = 2*( b-1) }}}
{{{ 10a + b }}} is the original number 
{{{ 10b + a }}} is the reverse of the number
{{{ 10b + a = 10a + b - 18 }}}
{{{ 10*(a-b) + b-a = 18 }}}
{{{ 10*(a-b) + b-a = 18 }}}
{{{ 10a - 10b + b - a = 18 }}}
{{{ 9a - 9b = 18 }}}
{{{ a - b = 2 }}}
By substitution:
{{{ 2*(b-1) - b = 2 }}}
{{{ 2b - 2 - b = 2 }}}
{{{ b = 4 }}}
and
{{{ a = 2*( b-1) }}}
{{{ a = 2*3 }}}
{{{ a = 6 }}}
The original number is 64
check:
Reverse the digits: 46
64 = 46 + 18 
OK