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I was reviewing rational inequalities, and noticed that the solution (identifying the domain of the inequality) was obtained just through critical points and a number line. Is there a way to get the domain using just reciprocals?

Example:

$$ \frac{x + 5}{x-4} \le 0 $$

I can see the crit points are -5 and 4, and the number line shows me that the interval is [-5,4). However, is there a way I can use the inequality itself and maybe reciprocals to derive the inequality in terms of x, such that: $$-5 \le x \lt 4$$

Thank you all for your time and help!

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  • $\begingroup$ $~\dfrac{A}{B} \leq 0~$ requires that $~B \neq 0,~$ or else the fraction is meaningless. Assuming that $~B~$ is prevented from ever equaling $~0,~$ then the inequality will hold if and only if one of the following is true: [1] $~A = 0.~$ [2] $~A > 0, ~B < 0.~$ [3] $~A < 0, ~B > 0.$ $\endgroup$ Mar 3, 2023 at 18:13

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\begin{align}&\frac{x + 5}{x-4} \le 0\\ \iff{}& 1+\frac{9}{x-4} \le 0\\ \iff{}& \frac{9}{x-4} \le-1\\ \color{violet}\iff{}& 0>\frac{x-4}{9} \ge-1\\ \iff{}& 0>{x-4} \ge-9\\ \iff{}& 4>x\ge-5.\end{align}

Regarding the third equivalence:

  1. $\color{violet}\Longrightarrow$ : the inequality on the left is negative, and on a negative domain the reciprocal function is a decreasing function, so applying the latter flips that inequality. Since a reciprocal has the same sign as the original expression, $\dfrac{x-4}9$ remains negative; it is necessary to make this explicit only so that we can say that...
  2. $\color{violet}\Longleftarrow$ : the rightmost inequality is negative, and on a negative domain the reciprocal function is a decreasing function, so applying the latter flips that inequality.
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