Question

Show work:
x/(x+1) < x/(x-1)

what I have so far:
x/(x+1) - x/(x-1)(x+1)(x-1)= 0(x+1)(x-1)... because you have to multiply both sides by the LCM which gives me:
x(x-1) - x(x+1) = 0 then x^2 - 1x - x^2 + 1x=0 which is 0 = 0 which doesnt work! what did i do wrong?

Answer 1

oops sorry, it should be x(x-1) - x(x+1) = 0 then x^2-1x-x^2-1x= 0 which is -2x=0 which is still 0 = 0

Answer 2

\[\frac{x}{(x+1)} < \frac{x}{(x-1)}\]
then:
\[\frac{x}{(x+1)} - \frac{x}{(x-1)}<0\]
then:
\[\frac{x(x-1)-x(x+1)}{(x+1)(x-1)}<0\]
then:
\[\frac{x^2-x-x^2-x}{(x+1)(x-1)}<0\]
then:
\[\frac{-2x}{(x+1)(x-1)}<0\]
Now you have to find the sign of the numerator and the sign of the denominator.
Then superpose them.
Please, tell me if you know how to continue.

Answer 3

shouldn't it be the equation > 0?

Answer 4

and cant i get rid of the (x+1) and (x-1) in the denominator when i multiply the equation by (x+1)(x-1) which is the LCM?

Answer 5

No, because on the first step we substract both sides x/(x-1).
And substracting doesn't make change in the sign.

Answer 6

No, you can't get rid of that when you do the LCM.

Answer 7

Alternatively, you can perform these steps:

1. Note that \(x \ne 1\)

\(\large\frac{x}{x+1} < \frac{x}{x-1}\)

2. Cross Multiply:
\(x(x-1) < x(x+1)\)

3. Distribute:
\(x^2 - x < x^2 + x\)

4. Subtract x^2 from both sides:
\(-x < + x\)

5. Add x to both sides
\(0 < 2x\)

6. Divide both sides by 2

\(x > 0\)

So we know that x > 0, except where x = 1

Answer 8

@Hero
We can't do the 2nd step:
We have to check that x+1 and x-1 are >0, Because if any of them are <0 we have to change the sign of the inequality.

Answer 9

once we solve for x, we'll know if it is greater than zero or not. Algebraically, cross-multiplication is a legal step.

Answer 10

In this case, x will not and cannot be less than or equal to zero

Answer 11

the answer is -1<x<0 and x>-1 i think

Answer 12

@Hero
I do not agree with that.
Here you can see that what you are saying is wrong:
http://www.wolframalpha.com/input/?i=x%2F%28x%2B1%29+%3C+x%2F%28x-1%29
The solution is
x>1
AND
-1<x<0

Answer 13

Yeah, I neglected to include the part where -1<x<0

Answer 14

oops correction to my comment -1<x<0 and x<1

Answer 15

Thanks guys! :)

Answer 16

@Elysse2012
Is there any doubt?
Be careful while working with this kind of problems.
Be careful with the "cross multiply"
I hope it helps!
:D

Answer 17

No doubt anymore! Thanks to the both of you!

Answer 18

You deserve that!
Thanks for the medal!
:D

Answer 19

My explanation was actually incomplete. If @radabc02, had not interrupted my explanation, I would have came to the same conclusion

Answer 20

What you do is note that \(x \ne -1\) and \(x \ne 1\). The zero is another boundary that should be noted. Doing so would give the boundaries:



Which after testing, we see that -1<x<0 and x>1