Question 481738
Here's the problem for reference:


{{{3/(x-9)=1/(x^2-9x)+2}}}



One thing to always remember: You CANNOT divide by zero. So something like {{{1/0}}} is undefined.





So this means that the denominators {{{x-9}}} and {{{x^2-9x}}} CANNOT be equal to zero (to avoid division by zero)



So if {{{x-9=0}}}, then {{{x=9}}}. This means that if x=9, then the first denominator is zero, which is not allowed.



Also, if {{{x^2-9x=0}}} then {{{x(x-9)=0}}} which means that {{{x=0}}} or {{{x=9}}} when we solve for x. So if x=0 or x=9, then the second denominator will be zero, which is not allowed.



So the following values of x will result in division by zero: 0, 9