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