FP2 inequalities question

\frac{2}{x-2}>\frac{1}{x+1}

So I worked it out and have got it down to two possible sets of values :

x>-4

-2<x<-1

x<-4

-1<x<-2

Now , my textbook says the answer should be the second option. But how is it possible that a number be greater than -2 and less than -1?( and be less than -4!)
Isn't the first option the only one possible?
( Im not allowed to solve graphically)

Reply 1

ghostwalker

Original post by princejan7

\frac{2}{x-2}>\frac{1}{x+1}

So I worked it out and have got it down to two possible sets of values :

x>-4

-2<x<-1

x<-4

-1<x<-2

Well the actual answer to what you've put is x > 2 or -4 < x < -1, which is neither of those. So, check your post and/or provide some working.

Reply 2

princejan7

oh, my mistake. :tongue:

So the two possible sets are

x>-4

2<x<-1

x<-4

-1<x<2

But I still dont understand why the second possibility is the answer..How can a number be between -1 and 2 and be less than -4?

Reply 3

marek35

Original post by princejan7

oh, my mistake. :tongue:

So the two possible sets are

x>-4

2<x<-1

x<-4

-1<x<2

But I still dont understand why the second possibility is the answer..How can a number be between -1 and 2 and be less than -4?

post your working.

What method are you using to solve these inequalities to check the regions?

Reply 4

gunmetalpanda

How? I get x>-4 :confused:

Reply 5

princejan7

Its kind of a long working and im not too familiar with Latex but the idea is if

\frac{x+4}{(x+1)(x-2)}<0

then either

Unparseable latex formula:

\((x+4)>0 , and (x+1)(x-2)<0

Unparseable latex formula:

\((x+4)<0 , and (x+1)(x-2)>0

Reply 6

ghostwalker

Original post by princejan7

Its kind of a long working and im not too familiar with Latex but the idea is if

\frac{x+4}{(x+1)(x-2)}<0

then either

Unparseable latex formula:

\((x+4)>0 , and (x+1)(x-2)<0

Unparseable latex formula:

\((x+4)<0 , and (x+1)(x-2)>0

I make it the other way around

\frac{x+4}{(x+1)(x-2)}>0

then either

(x+4)>0 , and (x+1)(x-2)>0

in which case x > -4 and [x < -1 or x >2] which leads to -4 < x < -1 or x > 2

or

(x+4)<0 , and (x+1)(x-2)<0

in which case x < -4 and -1 < x < 2 which has no solutions.

So the nett result is -4 < x < -1 or x > 2

(edited 13 years ago)

Reply 7

princejan7

Original post by ghostwalker

I make it the other way around

\frac{x+4}{(x+1)(x-2)}>0

then either

(x+4)>0 , and (x+1)(x-2)>0

in which case x > -4 and [x < -1 or x >2] which leads to -4 < x < -1 or x > 2

or

(x+4)<0 , and (x+1)(x-2)<0

in which case x < -4 and -1 < x < 2 which has no solutions.

So the nett result is -4 < x < -1 or x > 2

Why did you make it

\frac{x+4}{(x+1)(x-2)}>0

Could you please post a similar working with <0?

Because i cant seem to get the required answer of x<-4, -1<x<2

Reply 8

ghostwalker

Original post by princejan7

Why did you make it

\frac{x+4}{(x+1)(x-2)}>0

Post how you're getting it the other way around, then someone can check where you're making the slip.

Reply 9

princejan7

Original post by ghostwalker

Post how you're getting it the other way around, then someone can check where you're making the slip.[/QCould you please post a similar working with >0?

\frac{x+4}{(x+1)(x-2)}<0

x+4 > 0
x>-4

AND
(x+1)(x-2) < 0
x+1<0 x-2>0
x<-1 x>2
x+1>0 x-2<0
x>-1 x<2

So for this set , the result is x>-4 , -1<x<2

OR

x+4 < 0
x<-4

AND
(x+1)(x-2) > 0
x+1>0 x-2>0
x>-1 x>2
x+1<0 x-2<0
x<-1 x<2

And for this set, the result is x<-4 and x>2 or x<-1

The required answer of x<-4, -1<x<2 is a cross between them both.. :s-smilie:

Reply 10

ghostwalker

Original post by princejan7

\frac{x+4}{(x+1)(x-2)}<0

You've not said how you came to this part, and that is the bit that's wrong.

Reply 11

DFranklin

The correct inequality is as GW says:

\frac{x+4}{(x-2)(x+1)} > 0

Reply 12

princejan7

Original post by ghostwalker

You've not said how you came to this part, and that is the bit that's wrong.

Well, the original question was :

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

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

\frac{2(x+1)-(x-2)}{(x-2)(x+1)}<0

\frac{(x+4)}{(x-2)(x+1)}<0

Reply 13

ghostwalker

Original post by princejan7

Well, the original question was :

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

That's NOT what you put in your original post!

Reply 14

princejan7

Original post by ghostwalker

That's NOT what you put in your original post!

sorry!!

wish i had typed it out properly at the beginning... :facepalm:

Reply 15

ghostwalker

Original post by princejan7

sorry!!

wish i had typed it out properly at the beginning... :facepalm:

Me too!

Original post by princejan7

\frac{x+4}{(x+1)(x-2)}<0

x+4 > 0
x>-4

AND
(x+1)(x-2) < 0
x+1<0 x-2>0
x<-1 x>2
x+1>0 x-2<0
x>-1 x<2

So for this set , the result is x>-4 , -1<x<2

What you've done here, is say "if the numerator is positive", i.e. if x > -4, then x lies between -1 and 2.

So -1 < x < 2 is part of the solution, as that is the overlap between x > -4 and -1 <x <2.

OR

x+4 < 0
x<-4

AND
(x+1)(x-2) > 0
x+1>0 x-2>0
x>-1 x>2
x+1<0 x-2<0
x<-1 x<2

And for this set, the result is x<-4 and x>2 or x<-1

The required answer of x<-4, -1<x<2 is a cross between them both.. :s-smilie:

And here you have "if the numerator is negative", i.e. x < -4, then it is > 2 or < -1.

And the restriction becomes x < -4, as that is the overlap between the two intervals.

And hence your final solution is x < -4 or -1 < x < 2