# Solving an inequality

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

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

• Sep 10th 2010, 03:13 PM
Plato
There are three cases:

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

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

$x < -1\; \Rightarrow \;2x + 2 < x - 5$
• Sep 14th 2010, 04:04 AM
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.
• Sep 14th 2010, 05:37 AM
Quote:

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
• Sep 15th 2010, 12:14 AM
stupidguy
Quote:

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? (Wondering)
• Sep 15th 2010, 12:39 AM
Unknown008
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 AM
Quote:

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

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

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.
• Sep 15th 2010, 04:16 AM
stupidguy
Quote:

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?
• Sep 15th 2010, 07:06 AM
Quote:

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.
• Sep 15th 2010, 07:24 AM
stupidguy
Quote:

Quote:

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.
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. (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 PM
Krizalid
Quote:

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.
• Sep 16th 2010, 04:29 PM
Plato
I have been so aggravated by this thread. Here are the answers.
Quote:

Originally Posted by stupidguy
Solve the inequality $\frac{2}{x-5}<\frac{1}{x+1}$

The solution set is: $\left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$
Quote:

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 : $(-5,5)$.
• Sep 16th 2010, 06:04 PM
stupidguy
@mathaddict: I think there is a small correction required:

2x+2-x+5=x+7, not x+3.
• Sep 17th 2010, 03:41 PM
Plato
This was my original reply to you. You did nothing with it.
Quote:

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: $\left( { - \infty , - 7} \right) \cup \left( { - 1,5} \right)$
• Sep 17th 2010, 03:57 PM
Plato
Quote:

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.