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Bunuel
If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?




Attachment:
2A8vkix.jpg


|2z| – 1 ≥ 2

|2z| ≥ 3

|z| ≥ 3/2

So \(z \geq 1.5\) or \(z \leq -1.5\)

Answer (A)
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If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?




Attachment:
2A8vkix.jpg

Option A and Option D are same?
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Bunuel
If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?




Attachment:
2A8vkix.jpg

Option A and Option D are same?

No. Notice the bolded parts in those options. A reads: \(x \leq -1\) or \(x \geq 1\), while D reads \(-1 \leq x \leq 1\).
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Bunuel
If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?




Attachment:
2A8vkix.jpg

Option A and Option D are same?

No. Notice the bolded parts in those options. A reads: \(x \leq -1\) or \(x \geq 1\), while D reads \(-1 \leq x \leq 1\).

Thanks a lot Bunuel. Lots of love and blessings to you for helping all of us here on Gmatclub.
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Bunuel
If |2z| – 1 ≥ 2, which of the following graphs could be a number line representing all the possible values of z?


Attachment:
2A8vkix.jpg

IMPORTANT: This question is similar to "What must be true" questions. The difference is that the correct answer choice shows every possible solution (and no more) to be given inequality.

Let's TEST SOME values.

When I scan the answer choices, I see that some have 0 as part of the solution, and others don't.
So, let's see what happens when z = 0

Plug z = 0 into original equation to get: |2(0)| – 1 ≥ 2
Evaluate to get: |0| – 1 ≥ 2
We get: 0 – 1 ≥ 2
And then: -1 ≥ 2
NOT TRUE.
So, z = 0 is NOT a solution.
Therefore, we'll ELIMINATE D and E, since they say z=0 IS a solution.

Strategy: Among the three remaining answer choices, two have have 1 as part of the solution, and one doesn't.
So, let's see what happens when z = 1
Plug z = 1 into original equation to get: |2(1)| – 1 ≥ 2
Evaluate to get: |2| – 1 ≥ 2
We get: 2 – 1 ≥ 2
And then: 1 ≥ 2
NOT TRUE.
So, z = 1 is NOT a solution to the given inequality.
Therefore, we'll ELIMINATE B and C, since they say z = 1 IS a solution.

By the process of elimination, we're left with A, the correct answer.

Cheers,
Brent
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