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Re: If |2z| – 1 ≥ 2, which of the following graphs could be a nu [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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

Image


Another approach is to 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 include z=0 as a solution.

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.
Therefore, we'll ELIMINATE B and C, since they include z=1 as a solution.

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

Cheers,
Brent
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Re: If |2z| 1 2, which of the following graphs could be a nu [#permalink]
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Well, let´s see:
[2z] = 2z ou -2z, right?

Therefore,

2z -1 >= 2
z>= 3/2

OU

-2z -1 >= 2
-2z >= 3 multiplied by(-1)
2z <= -3
z <= -3/2

Option A
Piece of cake!
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Re: If |2z| 1 2, which of the following graphs could be a nu [#permalink]
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Re: If |2z| 1 2, which of the following graphs could be a nu [#permalink]
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