Question 480647
Let one be n
Let tens=t

{{{t+n=13}}} ----eq1
{{{10n+t=27+10t+n}}} -----eq2


Lets sorts the terms in eq2
{{{t-10t-n+10n=27}}}
{{{-9t+9n=27}}}


multiply eq1 by (9) 
{{{9t+9n=117}}}
{{{-9t+9n=27}}}
---------------
    {{{18n=144}}}
    {{{n=8}}}


substitute n=8 in eq1
So,
{{{t+8=13}}}
{{{t=5}}}


{{{t=5}}} and {{{n=8}}}
The original number is {{{5*10+8=50+8=58}}}