<s>[INST] Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then the trinomial is factorized
$$(x-3)(x-1) \le 0 $$
Therefore
$$1\lt x \lt 3 $$
Can someone tell me if I make a mistake, or if the process is good enough. Will you have done something different?
 [/INST] Never multiply an inequality by a variable since when multiplying by a negative number you have to change the sign.This is how this kind of problems should be solved
$$\frac{x^2+3}{x}\leq 4\\\frac{x^2+3}{x}-4\leq 0\\\frac{x^2-4x+3}{x}\leq 0\\\frac{(x-3)(x-1)}{x}\leq 0$$
Now you should spit it into cases


*

*$x<0$ Then you have that $x,(x-1),(x-3)$ are all negative so everything is negative

*$0<x<1$ Then you have that $x$ is positive and $(x-1),(x-3)$ negative hence the expression is positive since 2 negatives give positive

*$1<x<3$ Then you have that $x,(x-1)$ are positive and $x-3$ is negative hence the expression is negative

*$x>3$ everything is positive hence expression is positive


Now you see that $x<0$ and $1<x<3$ are solutions,now to check the boundaries,$x=0,1,3$ you can see that for $x=0$ it's undefined and that $x=1,3$ fit so the solution is $x\in(-\infty,0)\cup [1,3]$
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