$$\frac { x+2 }{ x+3 } <\frac { x-1 }{ x-2 } $$

This is what I got so far:

$$\frac { x+2 }{ x+3 } -\frac { x-1 }{ x-2 } <0$$

Now I am completely lost because I don't know the next step. This problem would take a combination of rules that I already learned to solve, but I don't know where to start.

I feel like the first thing to do would be to get a common denominator. However, that negative sign is really messing me up. Does it only apply to the numerator or both? Please point me in the right direction.


Finding the common denominator is spot on: $$\frac { x+2 }{ x+3 } -\frac { x-1 }{ x-2 } < 0 \iff \frac{ (x +2)(x - 2) - (x-1)(x+3)}{(x+3)(x-2)}\lt 0$$

$$\iff \frac{-1 - 2x}{(x+3)(x-2)} \lt 0 \iff (-1)\frac{1 + 2x}{(x+3)(x-2)}\lt 0$$ $$\iff \frac{1+2x}{(x+3)(x-2)} \gt 0$$

  • $\begingroup$ When you combined like terms in the second step... how did you exactly do that? $\endgroup$ – Cherry_Developer Aug 25 '14 at 20:16
  • $\begingroup$ I expanded each quadratic: $$(x+2)(x-2) - (x-1)(x+3) = x^2 - 4 -(x^2 + 2x-3) = x^2 - 4 -x^2 -2x+3 = -2x -1$$ $\endgroup$ – amWhy Aug 25 '14 at 20:22
  • $\begingroup$ I just experimented and got the same result. I also understand now that if multiplying a rational or fractional expression by (-1), only the numerator actually becomes negative. Is that correct? $\endgroup$ – Cherry_Developer Aug 25 '14 at 20:37
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    $\begingroup$ @Cherry_Developer In our system the equality holds no matter what. We can even have $(-1)\frac{x^{2014}\log(x^4+\cos(\theta))}{y^{2014}(x^{1012}+1)}=\frac{-x^{2014}\log(x^4+\cos(\theta))}{y^{2014}(x^{1012}+1)}=\frac{x^{2014}\log(x^4+\cos(\theta))}{-y^{2014}(x^{1012}+1)}\ge k$, if we wish. It doesn't matter in any case. $\endgroup$ – user26486 Aug 25 '14 at 20:47
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    $\begingroup$ @Cherry_Developer If we'd ended up with $\frac{1+2x}{-(x+3)(x-2)}<0$, yes. The answer would be exactly the same. $\endgroup$ – user26486 Aug 25 '14 at 20:59

We have$$\frac{x+2}{x+3}\lt\frac{x-1}{x-2}\iff 1-\frac{1}{x+3}\lt 1+\frac{1}{x-2}\iff -\frac{1}{x+3}\lt\frac{1}{x-2}.$$

If $x\lt -3$, then LHS is positive and RHS is negative.

If $x\gt 2$, then LHS is negative and RHS is positive.

If $-3\lt x\lt 2$, then multiplying the both sides by $(x+3)(x-2)\lt 0$ gives you $$-(x-2)\color{red}{\gt} x+3\iff x\lt -1/2.$$

Hence, the answer is $-3\lt x\lt -1/2$ or $x\gt 2$.


So, we have $$\frac{(x-2)(x+2)-(x-1)(x+3)}{(x+3)(x-2)}<0$$



Check separately for $x+3=0,x-2=0$

Otherwise, $$\frac{x+\dfrac12}{(x+3)(x-2)}>0\iff(x+3)(x-2)\left(x+\frac12\right)>0$$

  • $\begingroup$ you should also check for $x=-\frac{1}{2}$, as it implies a change of sign in the numerator $\endgroup$ – cjferes Aug 25 '14 at 16:21
  • $\begingroup$ @cjferes, I've multiplied by $$(x+3)^2(x-2)^2$$ SO, the sign of the last two expressions will be same if $$(x+3)(x-2)\ne0$$ $\endgroup$ – lab bhattacharjee Aug 25 '14 at 16:23
  • $\begingroup$ That's OK, your procedure is allright. I'm just noticing that he needs to verify changes of sign chaeking separately $x+3=0$, $x-2=0$, AND $x+\frac{1}{2}=0$ (you only missed the last condition). $\endgroup$ – cjferes Aug 25 '14 at 16:29
  • $\begingroup$ @cjferes He didn't miss anything. He said we needed to check $x+3=0$ and $x-2=0$ separately because only when $x\neq -3, x\neq 2$ do we have the fact that $\frac{x+\frac12}{(x+3)(x-2)}>0\iff(x+3)(x-2)\left(x+\frac12\right)>0$. After that, the reader is left to continue himself. $\endgroup$ – user26486 Aug 25 '14 at 21:09
  • $\begingroup$ didn't understand that, but i know his procedure is ok (also said that before). Thanks for the explanation $\endgroup$ – cjferes Aug 25 '14 at 21:18

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