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?
 A: In case you don't want to pass fraction(although it's probably harder).
$$\frac{x}{x-4}<\frac{x-5}{x+1}\iff x(x-4)(x+1)^2<(x-5)(x+1)(x-4)^2\iff $$
$$x(x-4)(x+1)^2-(x-5)(x+1)(x-4)^2<0\iff (x-4)(x+1)(x(x+1)-(x-5)(x-4))<0\iff$$
$$(x-4)(x+1)(10x-20)<0\iff(x-4)(x+1)(x-2)<0$$ use multiplicity and intermediate value theorem.
it works because you can multiply by squares, because you know squares are positive.
A: Just solve the equation $\frac{x}{x-4}=\frac{x-5}{x+1}$.  A change of sign can only occur at the zeroes of that equation, $x=2$, or where the equation is not defined i.e., at $x=4$ or $x-=-1$.  So check the signs on the correspondent intervals.
A: You can, as longer as you keep track of possible changes in the inequality's direction.
For example, you can say that for $x>4$, then $x-4>0$ and $x+1>0$ therefor you can find the solutions to:
$$(x+1)x<(x-5)(x-4)$$
(Answer $x<2$ but we assumed $x>4$ so the actual answer is none in this situation).
Also for $x<-1$ we have that $x+1<0$ and $x-4<0$, then we must solve:
$$(x+1)x<(x-5)(x-4)$$
(Yet again, the answer is $x<2$, which intersect with the assumption $x<-1$ for the real answer $x<-1$.)
The final case: for $-1<x<4$ we have $x+1>0$ and $x-4<0$ therefor we must solve
$$(x+1)x>(x-5)(x-4)$$
(And the answer is $x>2$, which combined to the assumption yields $2<x<4$)
So you can do it if you keep track of all possible changes in the signs due to negative factors, and then you remember to apply those conditions to the answer.
Or avoid altogether any multiplication by something that might be negative.
A: $$\frac x{x-4}<\frac{x-5}{x+1}\iff \frac x{x-4}-\frac{x-5}{x+1}<0\iff 10\frac{x-2}{(x-4)(x+1)}<0$$
$$\text{  Multiplying the numerator & the denominator by } \frac{(x-4)^2(x+1)^2}{10} \text{ which is}>0$$ 
$$\frac x{x-4}<\frac{x-5}{x+1}\iff (x-4)(x+1)(x-2)<0$$
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 
