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

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

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New post 07 Nov 2016, 23:59
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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

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T6774.png
T6774.png [ 2.39 KiB | Viewed 2302 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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New post 08 Nov 2016, 05:36
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Given: \(x \geq {-3}\) and \(x \leq {2}\)
\((x + 3) \geq {0}\) and \((x - 2) \leq {0}\)
\((x + 3) \geq {0}\) and \((2 - x) \geq {0}\)

Answer: 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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New post 08 Nov 2016, 06:53
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Bunuel wrote:
Image
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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New post 09 Nov 2016, 09:59
Bunuel wrote:
Image
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.

Answer: 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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New post 01 Feb 2019, 11:18
ScottTargetTestPrep wrote:
Bunuel wrote:
Image
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.

Answer: D


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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New post 04 Feb 2019, 06:54
ghnlrug wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
Image
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.

Answer: D


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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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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