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Is (x + 1)/(x - 3) < 0 ?

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Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 26 Mar 2016, 21:05
1
Is (x + 1)/(x - 3) < 0 ?

plotting the inequality we get the zone -1<x<3 satisfying the above inequality.

(1) -1 < x < 1
as this range is subset of -1<x<3 so the inequality will always be -ve. Sufficient.

(2) x^2 - 4 < 0

-2<x<2
For this range, we can have both +ve and -ve values for the given inequality. Insufficient.

Hence A.
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 03 Jul 2016, 00:02
Is (x + 1)/(x - 3) < 0 ?

(1) -1 < x < 1
(2) x^2 - 4 < 0

Can i use componendo dividendo here ?? (just checking as Im not sure)

(x + 1 + x -3 )/ (x + 1 - x + 3) < 0 (using a/b ===> (a + b)/(a - b)

(2x - 2)/(4) < 0

x/2 - 1/2 < 0

x/2 < 1/2 (adding 1/2 on both sides)

x < 1 (multiplying by 2 on both sides)

now take a look at the answer choices

1) -1 < x < 1 =====> clearly within our given value of x < 1 -----> hence ok

2) x^2 - 4 < 0
===> x^2 < 4
x < 2 or x > -2
so -2 > x > 2, if x is less than 1 than ok but if 1 < x < 2 then not ok

Hence ans is A)

Please help me out here? Need to verify this method before I applying it elsewhere
If its wrong, why is it wrong? Is it because we cannot apply componendo - dividendo to inequalities ?
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 13 Aug 2016, 05:53
Bunuel wrote:
Moved to DS subforum.

Lolaergasheva wrote:
Is x+1/x-3<0 ?

1) -1 < x < 1
2) x^2-4 < 0


Lolaergasheva, please format the questions properly:

Question should read:

Is (x+1)/(x-3)<0 ?

Is \(\frac{x+1}{x-3}<0\)? --> roots are -1 and 3, so 3 ranges: \(x<-1\), \(-1<x<3\) and \(x>3\) --> check extreme value: if \(x\) some very large number then \(\frac{x+1}{x-3}=\frac{positive}{positive}>0\) --> in the 3rd range expression is positive, then in 2nd it'll be negative and in 1st it'll be positive again: + - +. So, the range when the expression is negative is: \(-1<x<3\).

Thus the question basically becomes: is \(-1<x<3\)?

(1) -1 < x < 1. Sufficient.
(2) x^2-4 < 0 --> \(-2<x<2\). Not sufficient.

Answer: A.

Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486

And again:

!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.


Hello Bunuel,

Just confirming something -
1. When a negative number is moved and multiplied or divided on the other side of an inequality equation, only then will the inequality sign reverse.
2. The inequality sign does not reverse when a number of any sign is moved and multiplied or divided on the other side of the inequality TO A NEGATIVE number.

Please confirm if I am right in my understanding. Thanks.
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 14 Aug 2016, 00:06
TheLordCommander wrote:
Bunuel wrote:
Moved to DS subforum.

Lolaergasheva wrote:
Is x+1/x-3<0 ?

1) -1 < x < 1
2) x^2-4 < 0


Lolaergasheva, please format the questions properly:

Question should read:

Is (x+1)/(x-3)<0 ?

Is \(\frac{x+1}{x-3}<0\)? --> roots are -1 and 3, so 3 ranges: \(x<-1\), \(-1<x<3\) and \(x>3\) --> check extreme value: if \(x\) some very large number then \(\frac{x+1}{x-3}=\frac{positive}{positive}>0\) --> in the 3rd range expression is positive, then in 2nd it'll be negative and in 1st it'll be positive again: + - +. So, the range when the expression is negative is: \(-1<x<3\).

Thus the question basically becomes: is \(-1<x<3\)?

(1) -1 < x < 1. Sufficient.
(2) x^2-4 < 0 --> \(-2<x<2\). Not sufficient.

Answer: A.

Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486

And again:

!
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.


Hello Bunuel,

Just confirming something -
1. When a negative number is moved and multiplied or divided on the other side of an inequality equation, only then will the inequality sign reverse.
2. The inequality sign does not reverse when a number of any sign is moved and multiplied or divided on the other side of the inequality TO A NEGATIVE number.

Please confirm if I am right in my understanding. Thanks.


Adding or subtracting the same number on both sides of the inequality doesn't change the sign of the inequality.

Multiplying or dividing a -ve number reverses the sign of the inequality.

By these statements I also mean moving the number(from LHS to RHS).

Let me show you how :

x -2 > 0

We can say adding 2 on both sides is equivalent to moving - 2 to RHS.

or x-2+2 > +2

or x>2
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 29 Jan 2017, 05:31
is (x+1)/(x-3) < 0?

1) -1 < x < 1
2) x^2 - 4 < 0

So, let's see what the question is asking. Is the ratio of 2 quantities negative=> either the numerator OR the denominator must be negative for this to be true but not BOTH.

1) The numerator will be positive because x is greater than -1. The denominator will always be negative since the maximum x can get is very close to 1, but subtracting 3 from it, will result in a negative number. So, we have a negative result. SUFFICIENT.

2) X^2 <4 => -2<X<2. Now, the denominator, again, will be always negative since the MAX x can get is very close to 2. BUT, now the numerator can be positive [for x>-1] or negative [for -2< x<-1]. Hence, we can get a positive or a negative end result. INSUFFICIENT.

Hope this helps.

Hence A.
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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New post 29 Jan 2017, 05:37
is (x+1)/(x-3) < 0?

1) -1 < x < 1
2) x^2 - 4 < 0

So, let's see what the question is asking. Is the ratio of 2 quantities negative=> either the numerator OR the denominator must be negative for this to be true but not BOTH.

1) The numerator will be positive because x is greater than -1. The denominator will always be negative since the maximum x can get is very close to 1, but subtracting 3 from it, will result in a negative number. So, we have a negative result. SUFFICIENT.

2) X^2 <4 => -2<X<2. Now, the denominator, again, will be always negative since the MAX x can get is very close to 2. BUT, now the numerator can be positive [for x>-1] or negative [for -2< x<-1]. Hence, we can get a positive or a negative end result. INSUFFICIENT.

Hope this helps.

Hence A.
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Re: Is (x + 1)/(x - 3) < 0 ?  [#permalink]

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