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Current Student B
Joined: 24 Jul 2014
Posts: 21
Location: Germany
Concentration: Entrepreneurship, Technology
GMAT 1: 690 Q49 V35 GMAT 2: 720 Q49 V38 GPA: 3.8
WE: Information Technology (Computer Software)

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(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1811
Re: Inequation query  [#permalink]

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ir2pid wrote:
(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

I'd first multiply both sides by -1, so the two factors on the left side look the same; we can distribute the -1 through the '3-x' factor. Because we're multiplying by a negative, we must reverse the inequality:

(x+5)(x-3) < 0

Now we're multiplying two things, and getting a negative result, so one of the two things is positive, the other is negative. But x+5 is clearly larger than x-3, so it must be x+5 that is positive, and x-3 that is negative. If x+5 > 0, then x > -5, and if x-3 < 0, then x < 3. So -5 < x < 3.

I think you used the wrong inequality symbols in the parts I highlighted in red in the quoted text (they should both say "-5 < x < 3" ).
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Current Student B
Joined: 24 Jul 2014
Posts: 21
Location: Germany
Concentration: Entrepreneurship, Technology
GMAT 1: 690 Q49 V35 GMAT 2: 720 Q49 V38 GPA: 3.8
WE: Information Technology (Computer Software)

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IanStewart wrote:
ir2pid wrote:
(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

I'd first multiply both sides by -1, so the two factors on the left side look the same; we can distribute the -1 through the '3-x' factor. Because we're multiplying by a negative, we must reverse the inequality:

(x+5)(x-3) < 0

Now we're multiplying two things, and getting a negative result, so one of the two things is positive, the other is negative. But x+5 is clearly larger than x-3, so it must be x+5 that is positive, and x-3 that is negative. If x+5 > 0, then x > -5, and if x-3 < 0, then x < 3. So -5 < x < 3.

I think you used the wrong inequality symbols in the parts I highlighted in red in the quoted text (they should both say "-5 < x < 3" ).

The expression was given in the gmatclub tests solution for question ID:
Probability & Combinations :: M28-11

Guess it's an oversight

Thanks for the explanation, you made it really easy to follow.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: Inequation query  [#permalink]

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ir2pid wrote:
(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

Check out these posts:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/

They discuss how to handle various inequalities with multiple factors.
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Math Expert V
Joined: 02 Sep 2009
Posts: 58332
Re: Inequation query  [#permalink]

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ir2pid wrote:
IanStewart wrote:
ir2pid wrote:
(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

I'd first multiply both sides by -1, so the two factors on the left side look the same; we can distribute the -1 through the '3-x' factor. Because we're multiplying by a negative, we must reverse the inequality:

(x+5)(x-3) < 0

Now we're multiplying two things, and getting a negative result, so one of the two things is positive, the other is negative. But x+5 is clearly larger than x-3, so it must be x+5 that is positive, and x-3 that is negative. If x+5 > 0, then x > -5, and if x-3 < 0, then x < 3. So -5 < x < 3.

I think you used the wrong inequality symbols in the parts I highlighted in red in the quoted text (they should both say "-5 < x < 3" ).

The expression was given in the gmatclub tests solution for question ID:
Probability & Combinations :: M28-11

Guess it's an oversight

Thanks for the explanation, you made it really easy to follow.

GMAT Club's solution correctly says −5<x<3. I think you've just copied incorrectly.

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3074
Re: Inequation query  [#permalink]

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ir2pid wrote:
(x+5)(3−x)>0. This equation holds true for −5>x>3.

Without applying values for x and testing, what is the simplest way of getting the range of x as −5>x>3

Dear ir2pid

In questions like the one you posted, the Wavy Line Method is an elegant and quick way to find the range of x.

Here is what you do in the Wavy Line Method:

1. Express the given expression in standard factorized form, that is in the form:

$$(x-a)^{p}(x-b)^{q}$$ < or > 0

The inequality you presented above is already in factorized form.

But is it in Standard form? Not yet. [because one of the factors is (3-x). We need to convert it to the form (x-3)]

Let's multiply both sides of the inequality with -1. Note that multiplying both sides of an inequality with a negative number changes the sign of inequality. So, we get:

(x+5)(x-3) < 0

2. Plot the Zero Points on the number line. These are the points for which the factorized expression becomes equal to zero.

(x+5)(x-3) becomes equal to zero for x = -5 and x = 3. So, these are the zero points. Plot them on the number line.

3. Now, starting from the TOP RIGHT CORNER of the number line, draw a Wavy Line that passes through these zero points.

The parts where this wavy line is ABOVE the number line are the parts where the given expression will be positive.
The parts where this wavy line is BELOW the number line are the parts where the given expression will be negative.

That's it! From the Wavy Line we've drawn for your inequality, you can easily see that that the expression (x+5)(x-3) will be negative for -5 < x < 3

Practice the Wavy Line Method for one or two more questions and am sure you'll start finding questions like this one to be pretty easy. Hope this was useful!

Japinder
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Re: Inequation query  [#permalink]

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_________________ Re: Inequation query   [#permalink] 29 Jan 2018, 00:05
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