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I'm stuck at this seemingly simple problem again.

Solve the difference $$\frac{3x+1}{x+2}<2$$

I try to solve this in the intuitive way:

$$\frac{3x+1}{x+2}<2$$ $$=>3x+1<2(x+2)$$ $$=>3x+1<2x+4$$ $$=>x<3$$

I then read that the solution is $-2<x<3$

Where in the world did they get that $-2$ from?

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  • $\begingroup$ Make it homogeneous by moving $2$ to the left and simplifying to get: $\frac{x-3}{x+2}<0$. Now you see the zeros $-2$ and $3$. $\endgroup$
    – farruhota
    Jan 23 '19 at 18:17
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In this case you can't multiply both sides by $(x+2)$ as it might be negative for certain values of $x$ and that would require reversing the inequality sign. Try multiplying both sides by $(x+2)^2$ instead and see if the resulting quadratic leads anywhere.

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When you multiply by something negative, the inequality sign flips. So when you go from $$ \frac{3x+1}{x+2}<2 $$ to $$ 3x+1<2(x+2) $$ without changing the inequality sign, you're implicitly assuming that $x+2>0$. For $x+2<0$, we instead get $$ 3x+1>2(x+2) $$

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You should notice that you could only multiply both sides by $x+2$ and preserve the inequality sign as $3x+1<2(x+2)$ if and only if $x+2>0$. Otherwise, the inequality sign will change since you multiply both sides by a negative number

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