Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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

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

share|improve this answer
    
When you combined like terms in the second step... how did you exactly do that? –  Cherry_Developer Aug 25 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 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 at 20:37
1  
@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. –  mathh Aug 25 at 20:47
1  
@Cherry_Developer If we'd ended up with $\frac{1+2x}{-(x+3)(x-2)}<0$, yes. The answer would be exactly the same. –  mathh Aug 25 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 $x\gt 2$.

share|improve this answer

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

share|improve this answer
    
you should also check for $x=-\frac{1}{2}$, as it implies a change of sign in the numerator –  cjferes Aug 25 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 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 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. –  mathh Aug 25 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 at 21:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.