How do solve $\frac{3}{x - 2} \leq \frac{3}{x + 3}$ $3/(x-2)<=3/(x+3)$ algebraically?

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Answer 1 · 2016-10-05T20:42:29

$- 3 < x < 2$ $-3<x<2$

Slower way:

Bring everything to the right: $\frac{3}{x - 2} - \frac{3}{x + 3} \leq 0$ $3/(x-2) - 3/(x+3) \leq 0$
Lowest common denominator: $\frac{3 (x + 3) - 3 (x - 2)}{(x - 2) (x + 3)} \leq 0$ $(3(x+3)-3(x-2))/((x-2)(x+3))\leq 0$
Expand the numerator: $\frac{3 x + 9 - 3 x + 6}{(x - 2) (x + 3)} \leq 0$ $(3x+9-3x+6)/((x-2)(x+3))\leq 0$
$\frac{9 + 6}{(x - 2) (x + 3)} \leq 0$ $(9+6)/((x-2)(x+3))\leq 0$
$\frac{15}{(x - 2) (x + 3)} \leq 0$ $(15)/((x-2)(x+3))\leq 0$
Since $15$ $15$ is always positive, the sign of the fraction is decided by the sign of the denominator. $(x - 2) (x + 3)$ $(x-2)(x+3)$ represents a parabola with zeros in $x = - 3$ $x=-3$ and $x = 2$ $x=2$ . Such a parabola is negative between its solutions, as you can see in the graph here:
graph{(x-3)(x+2) [-4.96, 6.146, -2.933, 2.614]}

How do solve 3/(x-2)<=3/(x+3)3x−2≤3x+33/(x-2)<=3/(x+3) algebraically?