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$$\frac{x^2-3x-2}{x^2+5x+6}<\frac{2-x}{x^2-4}$$ First, I got the things it couldn't be, which were: -3, -2, and 2

I thought to factor everything I could, and I got $$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{2-x}{(x+2)(x-2)}$$ I also cancelled out one value on the right side $$\frac{x^2-3x-2}{(x+3)(x+2)}<\frac{-1}{x+2}$$ but now, I'm stuck.

Alex
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  • As long as you're cancelling things out, what about the $(x+2)$ terms on each side? – Ian Coley Sep 02 '13 at 19:19
  • Make a case distinction, $x < -3$, $-3 < x < -2$ and $-2 < x$, and multiply the inequality with $(x+3)(x+2)$. – Daniel Fischer Sep 02 '13 at 19:19
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    Bring the right hand term to the left and make a common denominator to begin with. – imranfat Sep 02 '13 at 19:20
  • Frank, be careful. If you multiply both sides by $x+2$, then you either change the direction of the inequality or not, depending on whether $x$ turns out to be greater or less than $2$. – G Tony Jacobs Sep 02 '13 at 19:21
  • Frank, you can't do that. Do you think that x² > x is the same as x > 1 ? After all I just cancelled an x – imranfat Sep 02 '13 at 19:23

2 Answers2

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You can add or subtract the same quantity from both sides of an inequality. In this case, I suggest adding $\frac{1}{x+2}$ to both sides, so you get $0$ on the right. You'll have two fractions on the left, which you need to combine by finding a common denominator. Then you'll have a single rational expression on the left, and you only need to determine which values of $x$ make it negative.

Does that all make sense?

G Tony Jacobs
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Starting from where you stopped, we have $$\dfrac{x^2-3x-2}{(x+3)(x+2)}+\dfrac{1}{x+2}<0$$ Adding, we have $$\dfrac{(x-1)^2}{(x+3)(x+2)}<0$$

As the numerator is non-negative, we can ignore it for now. So we need solutions to $$(x+3)(x+2)<0$$

You should be able to do this now simply by splitting the number line into three regions, viz. $x < -3, -3 < x < -2, x > -2$ and checking in each.

Dont forget to exclude $x=1$ if it is in the feasible region!

Macavity
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