Question 719778
A two-digit number is four times the sum of its digits. 


Let the number be 10x+y and x and y are whole numbers less than or equal to 9.
{{{10x+y=4(x+y)}}},
{{{10x+y=4x+4y}}}
{{{6x-3y=0}}}
{{{2x-y=0}}}



If its digits are reversed, the new number is 36 more than the original number. 
{{{10y+x=36+10x+y}}},
{{{10y-y+x-10x=36}}}
{{{-9x+9y=36}}}
{{{x-y=-4}}}


SYSTEM TO SOLVE:
{{{highlight(2x-y=0)}}}
{{{highlight(x-y=-4)}}}


Subtract second equation from the first equation.  x+0=4, {{{highlight(x=4)}}}.  Looks like {{{highlight(y=8)}}}.