the solution to

[sanju09]

∣(x + 1)/(x - 1)∣ < 4; x ≠ 1.

What is the solution to the above inequality?
(A) x < 3.5, x > 5.3
(B) x < 3/5, x > 5/3
(C) 3.5 < x < 5.3
(D) 3/5 < x < 5/3
(E) 0 < x < 3/5

[papgust]

-4 < x+1/x-1 < 4

-4x+4 < x+1 < 4x-4

-4x+3 < x < 4x-5

3/5 < x < 5/3

[sanju09]

-4 < x+1/x-1 < 4

-4x+4 < x+1 < 4x-4

-4x+3 < x < 4x-5

3/5 < x < 5/3

Are you positive that x - 1 is positive?

[rockeyb]

The answer is B .

x<3/5 and x> 5/3

|(x+1)/(x-1)| < 4.

Case 1 .

(x+1)/(x-1) < 4

x+1 < 4x - 4

5 < 3x

5/3 < x .


Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x < 3/5 .

[sanju09]

The answer is B .

x<3/5 and x> 5/3

|(x+1)/(x-1)| < 4.

Case 1 .

(x+1)/(x-1) < 4

x+1 < 4x - 4

5 < 3x

5/3 < x.


Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x < 3/5 .

Why can't we consider squaring both sides in this case?

https://www.beatthegmat.com/crucial-doub ... tml#235005

[spoiler]Only 1411 TIGERS LEFT!!!! SAVE TIGERS.

Have they bunged mating too?[/spoiler]

[papgust]

-4 < x+1/x-1 < 4

-4x+4 < x+1 < 4x-4

-4x+3 < x < 4x-5

3/5 < x < 5/3

Are you positive that x - 1 is positive?

I made a big mistake i guess. I intended to do rockey's approach but turned out something else on the screen. It's time to take some time off

[rockeyb]

The answer is B .

x<3/5 and x> 5/3

|(x+1)/(x-1)| < 4.

Case 1 .

(x+1)/(x-1) < 4

x+1 < 4x - 4

5 < 3x

5/3 < x.


Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x < 3/5 .

Why can't we consider squaring both sides in this case?

https://www.beatthegmat.com/crucial-doub ... tml#235005

I don't get it why do you want to square both sides of the inequality ?

[sanju09]

I don't get it why do you want to square both sides of the inequality ?

If you have already gone through the link that I have posted above, then you may realize that squaring both sides is the surest and shortest way of cracking such problems. It sails you out error free too, see what is in bold in your original post. Falling to the right answer by chance cannot be taken generally.

[rockeyb]

I don't get it why do you want to square both sides of the inequality ?

If you have already gone through the link that I have posted above, then you may realize that squaring both sides is the surest and shortest way of cracking such problems. It sails you out error free too, see what is in bold in your original post. Falling to the right answer by chance cannot be taken generally.

In the post here https://www.beatthegmat.com/crucial-doub ... tml#235005 you have given the answer to your own question .

this is an extract from your last post

Modulus Inequalities can be worked out by squaring both sides of the inequalities or by means of graphs. It is valuable to note that squaring both sides is legitimate only when both sides of the inequality are either zero or positive for all values of the variable(s) used, or else it is optional to work loose the inequalities using the graphical method.

As said we do not know the value of X if its +ve or -ve . The only thing we know that X can not be 1.

Now since we do not know what X stands for we have to consider 2 cases one +ve and other -ve . This is because what ever the value of X be (+ve or - ve ) the answer for mod of X will always be X only .

OR we can draw a graph and solve it as you have rightly said.

Lets consider a simpler example :

Q . what is the range of all possible values of |x-2| < 5 ?

As you can see we dont know the value of X here . So X can be any thing +ve or -ve .


Note that if you draw a number line we will have center of line at 2 and not 0 . And if you move 5 units either side of 2 you will have your two extremes that is 7 and -3 .

Same thing can be achieved by solving two cases

1.Right side of number line

x-2 < 5
x < 7

2. Left side of number line

-(x-2) < 5 OR x-2 > -5

x > -3

Now you get your two extremes .

Hope this makes sense .

[sanju09]

It is valuable to note that squaring both sides is legitimate only when both sides of the inequality are either zero or positive for all values of the variable(s) used, or else it is optional to work loose the inequalities using the graphical method

As said we do not know the value of X if its +ve or -ve . The only thing we know that X can not be 1.

Hope this makes sense .

I haven't said that the variable(s) must be non-negative for squaring, please read the bold part again.

Also see your following work

Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x < 3/5 .

Is there any placing mistake? Are you also positive that x - 1 is positive. Squaring both sides saves us from many such irregularities.

[rockeyb]

@sanju09

I see your point.

I forgot to flip the inequality sign .

So how do you suggest we solve this ? can u explain?

[sanju09]

@sanju09

I see your point.

I forgot to flip the inequality sign .

So how do you suggest we solve this ? can u explain?

sure

∣(x + 1)/(x - 1)∣ < 4

both sides squared

(x^2 + 2 x + 1)/ (x^2 - 2 x + 1) < 16

15 x^2 - 34 x + 15 > 0

(3 x - 5) (5 x - 3) > 0

[spoiler]x < 3/5, x > 5/3

B[/spoiler]

[rockeyb]

@sanju09

I see your point.

I forgot to flip the inequality sign .

So how do you suggest we solve this ? can u explain?

sure

∣(x + 1)/(x - 1)∣ < 4

both sides squared

(x^2 + 2 x + 1)/ (x^2 - 2 x + 1) < 16

15 x^2 - 34 x + 15 > 0

(3 x - 5) (5 x - 3) > 0

[spoiler]x < 3/5, x > 5/3

B[/spoiler]

But you too get the same answer .

the two roots

1. 3x -5 >0
3x >5
x>5/3

2. 5x -3 >0

5x > 3
x>3/5

So we get x > 3/5 and x > 5/ 3

which eventually mean x > 5/3 .

If you see the two cases I have presented :


Case 1 .

(x+1)/(x-1) < 4

x+1 < 4x - 4

5 < 3x

5/3 < x.


Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x > 3/5 .

we get the exact same result x > 5/3 and x> 3/5 .

And that is not what option B says .

Option B says x> 5/3 and x< 3/5

This can be a typo but none of the option seem to match in this case.

[sanju09]

@sanju09

I see your point.

I forgot to flip the inequality sign .

So how do you suggest we solve this ? can u explain?

sure

∣(x + 1)/(x - 1)∣ < 4

both sides squared

(x^2 + 2 x + 1)/ (x^2 - 2 x + 1) < 16

15 x^2 - 34 x + 15 > 0

(3 x - 5) (5 x - 3) > 0

[spoiler]x < 3/5, x > 5/3

B[/spoiler]

But you too get the same answer .

the two roots

1. 3x -5 >0
3x >5
x>5/3

2. 5x -3 >0

5x > 3
x>3/5

So we get x > 3/5 and x > 5/ 3

which eventually mean x > 5/3 .

If you see the two cases I have presented :


Case 1 .

(x+1)/(x-1) < 4

x+1 < 4x - 4

5 < 3x

5/3 < x.


Case 2 .

(x+1)/(x-1) > -4

x+1 > 4 - 4x .

-3 > -5x

x > 3/5 .

we get the exact same result x > 5/3 and x> 3/5 .

And that is not what option B says .

Option B says x> 5/3 and x< 3/5

This can be a typo but none of the option seem to match in this case.

Please take your time and revise all what you wrote in my contrast

[rockeyb]

I dont get it what am I missing here ?