4/x < -1/3, what is the range of x

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.

Please hit Kudos if you liked this answer

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!

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.

@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]

I am multiplying by 4 in denominators and
squaring both sides. But my answer is coming wrong. Why I can't do in such way.

KarishmaB Bunuel

Your cannot square an inequality until and unless you know the sign of both the sides.

Think about this:

3 < 4
Squaring: 9 < 16 (Valid)
When both sides are positive, we can square.


-4 < -3
Squaring: 16 < 9 (Not Valid)
But we can multiply both sides by -1 and flip the inequality sign to get 4 > 3. Now both sides are positive so we can square to get 16 > 9 (Valid)
When both sides are negative, we can square but we need to flip the inequality sign.

-4 < 3
Squaring: 16 < 9 (Not Valid)
We cannot square.

So only when the inequality has same sign on both sides and we know which sign it is, then can we square it.

Since we don't know what x is here, we cannot square.

How I solved in 40 seconds:

1) note that the two fractions are equal when x=-12, therefore because 4/x needs to be less than x=-12, the lower bound of x is -12. Any value of x less than -12, the inequality is false.
2) note that x cannot be positive, otherwise the inequality is false. Upper bound of X is 0.
Therefore -12<x<0, precisely D.

Hi KarishmaB,

If x<0, why did we not replace x with –x?

Edit: Got confused between modulus and inequality. Above query will be valid only if there was a related modulus in the question. Karishma has very kindly explained this point below.

If we are given that x < 0, the negative sign is already integrated in x. If we add another negative in front of it, it will become positive. Note that -x is not always a negative number. It will be a positive number if x < 0.

e.g. if x < 0, then say x = -5 (negative number)
Then -x = -(-5) = 5 (positive number)

Hi everyone,

Let me share a rule with respect to the inequalities:

Whenever there is a term in the denominator, we can remove it by multiplying both the sides with the square of that term.

Here, we can multiply both sides by the square of x.

12x < -x^2
X(x+12)<0