Question 643496
Typo, or trick question?
 
IF THE PROBLEM WAS REALLY
x-5/3-x<0 , which is the same as {{{x-5/3-x<0}}} ,
Then those {{{x}}} and {{{-x}}} cancel each other.
You are left with {{{-5/3<0}}} ,
which is true for any value of {{{x}}}
So if that was the problem, the answer is x = any real number.
 
I SUSPECT THE THE PROBLEM MAY HAVE BEEN
{{{(x-5)/(3-x)<0}}} , which is something different,
and can be written as (x-5)/(3-x)<0 .
In that case, the problem is not so simple.
Then, we have to think of different cases, and look at when the numerator and denomiantor are positive, negative or zero.
 
I consider the cases (x<3, x=3, x between 3 and 5, x-5, and x>5).
For each case, I figure out if {{{x-5}}}, the numerator of the rational expression {{{(x-5)/(3-x)}}} ,  is positive, negative, or zero.
I figure out if {{{3-x}}}, the denominator, is positive, negative, or zero.
From there, I figure out if the whole rational expression will be positive, negative, zero, or undefined. If numerator and denominator have the same sign (both positive, or both negative), dividing one by the other, I get a positive value for the rational expression . If numerator and denominator have opposite signs, then the rational expression will be negative.
To keep my thinking straight, I make myself a table like this:
{{{drawing(350,150, 0,60,0,30,
locate(1,7,(x-5)/(3-x)), locate(1.1, 14,3-x),
locate(1.1, 22,x-5),locate(3,29,x),
locate(13,5,"<0"), locate(13,14,">0"),
locate(13, 22,"<0"),locate(13,29,"<3"),
locate(20,5,undefined), locate(25,14,">0"),
locate(26, 22,0),locate(26,29,3),
locate(40,5,">0"),locate(40,14,">0"),
locate(40,22,">0"),locate(38,29,3<x<5),
locate(52,5,0),locate(52,14,0),
locate(51,22,">0"),locate(52,29,5),
locate(62,5,"<0"),locate(62,14,">0"),
locate(62,22,"<0"),locate(62,29,">5")
)}}}
 
MORE EXPLANATIONS:
(Feel free to skip them; I usually explain a lot more than needed, and bore people to tears).
 
LOOKING AT THE DENOMINATOR:
If {{{x=3}}} <---> {{{3-x=0}}} , then the rational expression {{{(x-5)/(3-x)}}} is undefined, because it has zero for a denominator. 
For {{{x<3}}}, then {{{3-x>0}}} and for {{{x>3}}}, then {{{3-x>0}}}.
 
LOOKING AT THE NUMERATOR:
For {{{x<5}}}, then {{{x-5<0}}}.
For {{{x=5}}}, then {{{x-5=0}}}.
For {{{x>5}}}, then {{{x-5>0}}}.
 
FOR THE WHOLE RATIONAL EXPRESSION:
Dividing (or multiplying) two expressions with opposite signs (one positive and one the other negative number), the result is negative,
If both expressions have the same sign (both positive, or both negative, their product or quotient is positive.
 
THE CASES:
We have to see what happens when {{{x}}} is less than 3, at 3, between 3 and 5, at 5, and when {{{x}}} is more than 5.
For {{{x<3}}} , {{{3-x>0}}}, while {{{x<5}}} , and {{{x-5<0}}}.
In that case {{{(x-5)/(3-x)<0}}} , telling us that {{{x<3}}} is part of the solution
For {{{x=3}}}, the expression {{{(x-5)/(3-x)}}} does not exist.
For {{{x}}} between 3 and 5, nummerator and denominator are both negative,
so {{{(x-5)/(3-x)>0}}} .
For {{{x=5}}} , {{{x-5=0}}} and {{{(x-5)/(3-x)=0}}}
For {{{x>5>3}}} , {{{3-x<0}}}, while {{{x-5>0}}} , so {{{3-x<0}}}, so {{{x-5>0}}} is negative again.
In that case ({{{(x-5)/(3-x)<0}}}), telling us that {{{x>5}}} is part of the solution.
The solution is {{{x<3}}} or {{{x>5}}} .