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VeritasKarishma apologies for bothering you. I read your inequality series and it's been enlightening.

I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".

Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.

Would that work? Thank you.
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VeritasKarishma apologies for bothering you. I read your inequality series and it's been enlightening.

I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".

Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.

Would that work? Thank you.


You don't need to make it so complicated though what you suggest works too. The reason why factors in the denominator work is simply this:

When is abcd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

When is ab/cd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

Aren't the two cases the same? (Except that c and d cannot be 0 in the second case) Anyway, since the expression needs to be positive, a, b, c and d cannot be 0 in first case either.

Does it matter whether c and d are in numerator or denominator? No.
The signs of a, b, c and d decide the sign of the expression irrespective of whether they are in numerator or denominator. The case is the same here.
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We have the values of x as -3,-1,2 and 5.

So, substituting these values of the number line, we will the range of x as
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Wavy Line Method Application - Exercise Question #4


Find the range of values of x that satisfy the inequality \(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)




Wavy Line Method Application has been explained in detail in the following post:: https://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html


Detailed solution will be posted soon.

Equation is \(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)

We have the values of x as -3,-1,2 and 5.

Since x cannot be equal to 5 or -3 we need to exclude those values(they are at the denominator and we cannot have 0 at the denominator)

So, substituting these values of the number line, we will the range of x as

(-infinity,-3) U (-1,2) U (5,infinity)

Please correct me if I am missing anything.








Hello! Can you please explain how did you get x < -3? The eqn = x -5 > 0 i.e. x >5. So how come for x+3 > 0 it becomes x<-3? Shouldnt it be x>-3?
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Solution:

Hey Everyone,

Please find below the solution of the given problem,

Plotting the zero points and drawing the wavy line:



Required Range:

x < -3 or -1 < x < 2 or x > 5

Correct Answer: Option C


Where can i get the wavy ine method?
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Solution:

Hey Everyone,

Please find below the solution of the given problem,

Plotting the zero points and drawing the wavy line:



Required Range:

x < -3 or -1 < x < 2 or x > 5

Correct Answer: Option C


Where can i get the wavy ine method?

Check this post: https://gmatclub.com/forum/inequalities ... 91482.html
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