Method 1:

\(\frac{4}{x}<-\frac{1}{3}\)

\(\frac{4}{x} + \frac{1}{3} < 0\)

\(\frac{x + 12}{3x} < 0\)

Using the wavy line method discussed here: https://www.veritasprep.com/blog/2012/0 ... e-factors/

-12 < x < 0

Answer (D)

Method 2:

\(\frac{4}{x}<-\frac{1}{3}\)

x can be positive, or negative so we need to take two cases:

Case 1: If x > 0, multiply both sides by 3x without flipping the inequality sign

\(12 < -x\)

Multiply both sides by -1 (this will flip the inequality sign)

\(x < -12\)
But x was assumed to be non negative here so we have no solution in this range.

Case 2: If x < 0, multiply both sides by 3x by flipping the inequality sign.

\(12 > -x\)

Multiply both sides by -1 (this will flip the inequality sign)

\(x > -12\)

So here, -12 < x < 0

Answer (D)
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From the question stem we can conclude that x<0.

If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):

4>\(-(\frac{1}{3})x\)

From here we can conclude that the maximum value x can be is -12, in order to make the inequality true.

Hence -12<x<0.

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From the question stem we can conclude that x<0.

If we move x to the other side of the inequality, it changes sign (as we multiply with a negative):

4>\(-(\frac{1}{3})x\)

From here we can conclude that the maximum value x can be must be smaller than -12, in order to make the inequality true.

Hence -12<x<0.

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Hi VeritasKarishma,
I have doubt in second method (excuse me if very silly and basic).

Please explain that how could we conclude :
-12 < x < 0
from the information that: x > -12.

Regards,
Tamal

can anyone draw a wavy curvy line for ths expression ? -12 < x < 0


\(\frac{x + 12}{3x} < 0\) in nummerator x = - 12 but what do with 3x in the denominator ?

hello generis is my understanding correct ? took me a while to figure out how curvy method works

\(\frac{x+12}{3x} <0\)

To solve rational inequality:

- solve for when numerator equals 0

- find the values that make denominator equal 0


\(x+12=0\)
\(x = -12\)

\(3*0=0\)
\(x=0\)

So the critical points of boundaries are -12 and 0. Now to find the range draw a number line.




- = \(x<0\)

+ = \(x>0\)

We need to find X less than ZERO, so I marked that range.

Hey tamal99

Eventough I am not Karishma I will try to answer your question.

In the second Case marked above we are trying to find values that are below zero (x<0), therefore x cannot be greater than 0! Maybe picking values makes it clearer:

Try to plug in numbers above 0. \(\frac{4}{x}<-\frac{1}{3}\) No positive value will make this inequality true.

Therefore the range is -12<x<0.

Hope it makes sense!
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Hi tamal99

i think we need to set denominator equal zero , pls check the link below

https://study.com/academy/lesson/solvin ... ities.html

hopefully i answered your question correctly

In case 2, we have assumed that x is negative (hence have flipped the inequality sign later)
From the inequality, we got x > -12.
So x is negative but greater than -12. This gives us -12 < x < 0.
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@Bunuel can you help advise how to draw the below equation on the number line with +/- sections as explained above.

\(\frac{x + 12}{3x} < 0\) in nummerator x = - 12 but what do with 3x in the denominator ? [/quote]