Question 552895
Let x be the tens digit, and y the ones digit
The number's value is
{{{10x+y}}}, while the value of the number with the digits reversed is
{{{10y+x}}}
We know that {{{x+y=15}}} and that
{{{10y+x=(10x+y)+9}}} --> {{{10y+x=10x+y+9}}} --> {{{9y-9x=9}}} --> {{{9(y-x)=9}}}  {{{y-x=1}}}
We solve the system
{{{x+y=15}}}
{{{-x+y=1}}}
any way we can to find {{{y=8}}} and{{{x=7}}}
So the original number is 78.