$\frac{\left(x+1\right)}{x-2}-\frac{\left(x+3\right)}{x-4} \geq 0 \Leftrightarrow \frac{(x+1)(x-4)-(x+3)(x-2)}{(x-2)(x-4)}=\frac{-2(2x-1)}{(x-2)(x-4)}=\frac{-2(x-2)(2x-1)(x-4)}{(x-2)^2(x-4)^2} \geq 0 \Rightarrow (x-2)(2x-1)(x-4) \leq 0.$
This gave me the solutions $x_1 \leq 2,\; x_2 \leq 4,\; x_3 \leq \frac{1}{2}.$ After having controlled those with the original inequality, I got the final solution $(2,4) \cup(-\infty, \frac{1}{2}].$
It might just be me, but should not the solutions I get be automatically true for the inequality I try to solve, except for false roots et cetera? Or am I just babbling about something at the moment because I do not really understand what it is that I am talking about? Or did I actually do something wrong while solving the inequality?
Please help me get this straight.
Your solution is right, but I think it's better from $$\frac{-2(2x-1)}{(x-2)(x-4)}\geq0$$ to write the answer $(-\infty,\frac{1}{2}]\cup(2,4)$ immediately by the intervals method.
The intervals method it's the following.
You need to draw the x-axis and to put there points: $\frac{1}{2}$ (a full point because $\frac{1}{2}$ is one of solutions),
$2$ and $4$ (empty points).
Easy to see that on $(4,+\infty)$ our expression is negative.
Now, we get the following signs of the intervals (from the left to the right): $+,-,+,-$ and we got the answer.
The rule for choosing of signs is the following.
$n$ is called a degree of $a$ if we have an expression $(x-a)^n$.
Easy to understand that:
We see that in the expression $\frac{-2(2x-1)}{(x-2)(x-4)}$ degrees of the points $2$, $4$ and $\frac{1}{2}$ they are odds because $$\frac{-2(2x-1)}{(x-2)(x-4)}=\frac{-2(2x-1)^1}{(x-2)^1(x-4)^1},$$ which says that all signs are changed and we got the answer.
Another example.
Let, we need to solve $$\frac{(x-2)^2}{(x-1)(x-3)}\geq0.$$ The series of the signs is $+$, $-$, $-$, $+$ (draw it!) and we got the answer: $$(-\infty,1)\cup(3,+\infty)\cup\{2\}$$