Question 468789
Let x is the tens digit
Let y is the ones digit
Then number is {{{10x+y}}}
the tens digit is 5 less than the ones digit, then {{{y-x=5}}}.
The number is 14 greater than the product of the digits, then {{{10x+y-xy=14}}}
The system of equations
{{{system(y-x=5,10x+y-xy=14)}}}
Use the substitution method
From the first equation {{{y=5+x}}}
{{{10x+(5+x)-x(5+x)=14}}}
{{{10x+5+x-5x-x^2-14=0}}}
{{{x^2-6x+9=0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-(-6) +- sqrt( (-6)^2-4*1*9 ))/(2*1) }}} 
{{{x = (6 +- sqrt( 0))/2 }}} 
{{{x=3}}} the tens digit
{{{y=5+3=8}}} the ones digit
The number is 38