Carcass wrote:

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

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 = 0Plug

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 = 1Plug

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

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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