<s>[INST] Inequalities In Algebra So the problems ask to find Find all values of $x$ for which 
$$\frac{x}{x-4}<\frac{x-5}{x+1}.$$
So the solution requires moving both fractions to one side, finding a common denominator, combining, then factoring. Then you have a set of possible solutions for $x$ and you go on from there. 
(If anyone is interested, the solution is $x < -1$, $2 < x < 4$.)
Why can I simply multiply both side by $(x+1)$ and $(x-4)$ WITHOUT shifting everything to one side? 
I suspect it's because I don't really know if $(x-4)$ and $(x+1)$ are positive or negative. Since I don't know, I cannot appropriately flip the inequality. Hence, a tactic applicable for equalities simply doesn't work in this case.
Is it really just that or is there's a more fundamental reason?
 [/INST] $$\frac x{x-4}<\frac{x-5}{x+1}\iff \frac x{x-4}-\frac{x-5}{x+1}<0\iff 10\frac{x-2}{(x-4)(x+1)}<0$$
$$\text{  Multiplying the numerator & the denominator by } \frac{(x-4)^2(x+1)^2}{10} \text{ which is}>0$$ 
$$\frac x{x-4}<\frac{x-5}{x+1}\iff (x-4)(x+1)(x-2)<0$$
so we need odd number of factors $<0$
If all of the three are $<0$ we need $x<$min$(4,-1,2)=-1$
If $x-4<0\iff x<4,$ we need $x>$max $(-1,2)=2\implies 2<x<4$
If $x+1<0\iff x<-1,$ we need $x>$max $(4,2)=4$ which is impossible 
If $x-2<2\iff x<2,$ we need $x>$max $(4,-1)=4 $ which is impossible 
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