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# The entire range of values of x is marked by the dark region on the nu

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Math Expert
Joined: 02 Sep 2009
Posts: 61243
The entire range of values of x is marked by the dark region on the nu  [#permalink]

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07 Nov 2016, 23:59
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Difficulty:

45% (medium)

Question Stats:

66% (01:47) correct 34% (01:42) wrong based on 268 sessions

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The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:

T6774.png [ 2.39 KiB | Viewed 3003 times ]

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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08 Nov 2016, 05:36
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2
Given: $$x \geq {-3}$$ and $$x \leq {2}$$
$$(x + 3) \geq {0}$$ and $$(x - 2) \leq {0}$$
$$(x + 3) \geq {0}$$ and $$(2 - x) \geq {0}$$

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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08 Nov 2016, 06:53
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Top Contributor
Bunuel wrote:

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

One option is to TEST some values.

According to the diagram, 0 IS a solution to the inequality. Let's test each answer choice to see whether 0 is a solution to the given inequality. If it is NOT a solution, we'll eliminate that answer choice.
A. (3+0)(2−0) = 6 BUT it is NOT the case that 6 ≤ 0 ELIMINATE A
B. (3−0)(2+0) = 6 AND it IS the case that 6 ≥ 0 KEEP B
C. (0−3)(0+2) = -6 AND it IS the case that -6 ≤ 0 KEEP C
D. (0+3)(2−0) = 6 AND it IS the case that 6 ≥ 0 KEEP D
E. (3−0)(2−0) = 6 AND it IS the case that ≥ 0 KEEP E

Test another value. According to the diagram, 3 is NOT a solution to the inequality. Let's test each remaining answer choice to see whether 3 is a solution to the given inequality. If it IS a solution, we'll eliminate it.
B. (3−3)(2+3) = 0 and it IS the case that 0 ≥ 0. So ELIMINATE B
C. (3−3)(3+2) = 0 and it IS the case that 0 ≤ 0. So ELIMINATE C
D. (3+3)(2−3) = -6 and it is NOT the case that -6 ≥ 0 KEEP D
E. (3−3)(2−3) = 0 and it IS the case that 0 ≥ 0. So ELIMINATE E

By the process of elimination, the correct answer is D

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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09 Nov 2016, 09:59
Bunuel wrote:

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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01 Feb 2019, 11:18
ScottTargetTestPrep wrote:
Bunuel wrote:

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

Dear ScottTargetTestPrep,

In the last step, did you simply then multiply the two inequalities? Or how did you arrive at the conclusion that the only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0?
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Re: The entire range of values of x is marked by the dark region on the nu  [#permalink]

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04 Feb 2019, 06:54
ghnlrug wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0

Attachment:
T6774.png

Looking at the given number line, we see that the dark region on the number line represents all the values of x between -3 and 2, inclusive. In other words, it is -3 ≤ x ≤ 2. Thus:

-3 ≤ x AND x ≤ 2.

Manipulating each inequality we have:

-3 ≤ x

x ≥ -3

x + 3 ≥ 0

x ≤ 2

0 ≤ 2 - x

2 - x ≥ 0

x + 3 ≥ 0 AND 2 - x ≥ 0

The only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0.

Dear ScottTargetTestPrep,

In the last step, did you simply then multiply the two inequalities? Or how did you arrive at the conclusion that the only answer choice that will satisfy both of these inequalities is (x+3)(2−x) ≥ 0?

Yes, I multiplied the inequalities.
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# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
181 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: The entire range of values of x is marked by the dark region on the nu   [#permalink] 04 Feb 2019, 06:54
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