1. ## Solving an inequality

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

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

2. There are three cases:

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

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

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

3. 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.

4. Originally Posted by stupidguy
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.
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.

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

$\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 $|x|\geq 0$

so the only solution would be |x|<5 or -5<x<5

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.

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

$\frac{x+3}{(x-5)(x+1)}<0
$

The solution set is {x: x<-3 , -1<x<5}

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

then the solutions are a<-3 and -1<a<5

back substituting, |x|<-3 and -1<|x|<5

Obviously, |x|<-3 and |x|>-1 are not valid since $|x|\geq 0$

so the only solution would be |x|<5 or -5<x<5
Why cant you cross multiply the denominator?

Why cant |x|>-1 since |-3|>-1. |3|>-1, |0|>-1?

6. 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.

7. Originally Posted by stupidguy
Why cant you cross multiply the denominator?

Why cant |x|>-1 since |-3|>-1. |3|>-1, |0|>-1?
Perhaps i was a bit unclear, 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.

Perhaps i was a bit unclear, 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.
so can u change accordingly to make it clearer?

9. Originally Posted by stupidguy
so can u change accordingly to make it clearer?

|x|>-1 is always true for all $x\in R$ so there will be no restriction on x here and the inequality will not be affected.

Perhaps i was a bit unclear
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.
unclear $\neq$ wrong -.-

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

But in the end I still benefit from your help.

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.....

11. Originally Posted by stupidguy
another qn:
if |x|>-0.05, there is infinite solutions, hence no solution, right?
if |x|<-0.05, there is also no solution, right?
first one has infinite solutions, but not by having them we can say that there's no solution, it's stupid. xD

and second one doesn't have solution.

12. I have been so aggravated by this thread. Here are the answers.
Originally Posted by stupidguy
Solve the inequality $\frac{2}{x-5}<\frac{1}{x+1}$
The solution set is: $\left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$
Originally Posted by stupidguy
Solve the inequality $\frac{2}{x-5}<\frac{1}{x+1}$
Hence, solve $\frac{2}{|x|-5}<\frac{1}{|x|+1}$
The solution set is : $(-5,5)$.

13. @mathaddict: I think there is a small correction required:

2x+2-x+5=x+7, not x+3.

14. This was my original reply to you. You did nothing with it.
Originally Posted by Plato
There are three cases:
$x > 5\; \Rightarrow \;2x + 2 < x - 5$
$- 1 < x < 5\; \Rightarrow \;2x + 2 > x - 5$
$x < -1\; \Rightarrow \;2x + 2 < x - 5$
Take case i) $x<-7$ so the intersection is empty.
For case ii) $x>-7$ so the intersection is $-1.
For case iii) $x<-7$ so the intersection is $x<-7$.

Unite those set and you get: $\left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$

15. Originally Posted by permamentlybannedguy
can u use the conventional way like wat mathaddict did and continue there......i seriously dun understand what are u doing. in the 1st place, why did u cross multiply them?
I am truly sorry that you do not find that helpful.
I am a mathematician, so I do mathematics.
I am sure that that there are many other ways to do this problem.