# Stuck solving a nonlinear inequality

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

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

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

-
When you combined like terms in the second step... how did you exactly do that? –  Cherry_Developer Aug 25 '14 at 20:16
I expanded each quadratic: $$(x+2)(x-2) - (x-1)(x+3) = x^2 - 4 -(x^2 + 2x-3) = x^2 - 4 -x^2 -2x+3 = -2x -1$$ –  amWhy Aug 25 '14 at 20:22
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? –  Cherry_Developer Aug 25 '14 at 20:37
@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. –  user31415 Aug 25 '14 at 20:47
@Cherry_Developer If we'd ended up with $\frac{1+2x}{-(x+3)(x-2)}<0$, yes. The answer would be exactly the same. –  user31415 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 $$\frac{(x-2)(x+2)-(x-1)(x+3)}{(x+3)(x-2)}<0$$

$$\iff\frac{(x^2-4)-(x^2+2x-3)}{(x+3)(x-2)}<0$$

$$\iff-\frac{2x+1}{(x+3)(x-2)}<0\iff\frac{x+\dfrac12}{(x+3)(x-2)}>0$$

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

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

-
you should also check for $x=-\frac{1}{2}$, as it implies a change of sign in the numerator –  cjferes Aug 25 '14 at 16:21
@cjferes, I've multiplied by $$(x+3)^2(x-2)^2$$ SO, the sign of the last two expressions will be same if $$(x+3)(x-2)\ne0$$ –  lab bhattacharjee Aug 25 '14 at 16:23
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). –  cjferes Aug 25 '14 at 16:29
@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. –  user31415 Aug 25 '14 at 21:09
didn't understand that, but i know his procedure is ok (also said that before). Thanks for the explanation –  cjferes Aug 25 '14 at 21:18