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$\frac{\left(x+1\right)}{x-2}-\frac{\left(x+3\right)}{x-4} \geq 0 \Leftrightarrow \frac{(x+1)(x-4)-(x+3)(x-2)}{(x-2)(x-4)}=\frac{-2(2x-1)}{(x-2)(x-4)}=\frac{-2(x-2)(2x-1)(x-4)}{(x-2)^2(x-4)^2} \geq 0 \Rightarrow (x-2)(2x-1)(x-4) \leq 0.$

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.

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    $\begingroup$ You found 3 solutions that satisfy the equality. Then you have said that each factor is less than zero. Then you proceed to find how each of the factor inequalities relate to the global inequality - and found your solution, what is wrong. $\endgroup$ – Chinny84 Sep 6 '18 at 11:46
  • $\begingroup$ why did u make it so hard? I meant from the 3th part to 4th? How did you that? Are you sure what you multiplied both nom. and den. of the 3th fraction is positive? $\endgroup$ – mrs Sep 6 '18 at 11:46
  • $\begingroup$ @mrs it is because the denominator is strictly positive - I wondered about that for a second. $\endgroup$ – Chinny84 Sep 6 '18 at 11:47
  • $\begingroup$ @Chinny84, it might just be me who am new to mathematics, in some sense. $\endgroup$ – ಠ ಠ Sep 6 '18 at 11:49
  • $\begingroup$ @ಠಠ: Yet, I have not been convinced why you did it like, however; I understand the job you did in 4th term. $\endgroup$ – mrs Sep 6 '18 at 12:00
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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:

if we pass through a point $a$ with even degree then the sign of the full expression does not change.

if we pass through a point $a$ with odd degree then the sign of the full expression is changed.

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: $$(-\infty,1)\cup(3,+\infty)\cup\{2\}$$

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  • $\begingroup$ What do you mean by the "intervals method"? $\endgroup$ – ಠ ಠ Sep 6 '18 at 14:11
  • $\begingroup$ @ಠ ಠ I added something. See now. $\endgroup$ – Michael Rozenberg Sep 6 '18 at 15:10
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I have another way which is an illustrating good one for you. I started from @Michael point. Try to find what values of $x$, make that fraction negative. Be careful of $x=2$ and $x=4$. Look at this graphic table. I hope you find the whole story which both of us are trying to tell you:

enter image description here

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