Solve the inequality $\displaystyle \frac{2}{x-5}<\frac{1}{x+1}$

Hence, solve $\displaystyle \frac{2}{|x|-5}<\frac{1}{|x|+1}$

(Headbang)

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- Sep 10th 2010, 02:56 PMstupidguySolving an inequality
Solve the inequality $\displaystyle \frac{2}{x-5}<\frac{1}{x+1}$

Hence, solve $\displaystyle \frac{2}{|x|-5}<\frac{1}{|x|+1}$

(Headbang) - Sep 10th 2010, 03:13 PMPlato
There are three cases:

$\displaystyle x > 5\; \Rightarrow \;2x + 2 < x - 5$

$\displaystyle - 1 < x < 5\; \Rightarrow \;2x + 2 > x - 5$

$\displaystyle x < -1\; \Rightarrow \;2x + 2 < x - 5$ - Sep 14th 2010, 04:04 AMstupidguy
how to solve part a? can you provide step by step guide? if not, any resource I can refer to cos I totally dun understand the lesson.

- Sep 14th 2010, 05:37 AMmathaddict
For part(a), you should know you can't multiply either of the denominators like how an algebraic equation involving rational function is solved. First step, move all function to one side and simplify.

$\displaystyle \frac{2}{x-5}-\frac{1}{x+1}<0$

$\displaystyle \frac{x+7}{(x-5)(x+1)}<0

$

The solution set is {x: x<-7 , -1<x<5}

As for part (b), let |x|=a

then the solutions are a<-7 and -1<a<5

back substituting, |x|<-7 and -1<|x|<5

Obviously, |x|<-7 and |x|>-1 are not valid since $\displaystyle |x|\geq 0$

so the only solution would be |x|<5 or -5<x<5 - Sep 15th 2010, 12:14 AMstupidguy
- Sep 15th 2010, 12:39 AMUnknown008
You cannot cross multiply because x can take any value, which includes values less for example less than 1. If that happened, then you'll be multiplying everything by a negative value, which will change the sign of the inequality. Since you don't know the value of x, you mustn't multiply.

- Sep 15th 2010, 03:22 AMmathaddict
- Sep 15th 2010, 04:16 AMstupidguy
- Sep 15th 2010, 07:06 AMmathaddict
- Sep 15th 2010, 07:24 AMstupidguyQuote:

Perhaps i was a bit unclear

Quote:

it's not that |x| cannot be greater than -1. In fact, it's always greater than -1 regardless of the value of x so it will not affect the inequality.

I tot you were unclear so I keep thinking |x|>-1 is not valid since because $\displaystyle x|\geq 0$ which doesn't make sense. (Punch)(Headbang)

But in the end I still benefit from your help. (Bow)(Clapping)

another qn:

if |x|>-0.05, there is infinite solutions, hence no solution, right?

if |x|<-0.05, there is also no solution, right?

confirming..... - Sep 16th 2010, 01:30 PMKrizalid
- Sep 16th 2010, 04:29 PMPlato
- Sep 16th 2010, 06:04 PMstupidguy
@mathaddict: I think there is a small correction required:

2x+2-x+5=x+7, not x+3. - Sep 17th 2010, 03:41 PMPlato
This was my original reply to you.

**You did nothing with it**.

Take case i) $\displaystyle x<-7$ so the intersection is empty.

For case ii) $\displaystyle x>-7$ so the intersection is $\displaystyle -1<x<5$.

For case iii) $\displaystyle x<-7$ so the intersection is $\displaystyle x<-7$.

Unite those set and you get: $\displaystyle \left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$ - Sep 17th 2010, 03:57 PMPlato