Answer: 0 and 1
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Work Shown:
[tex]|2x - 1| <3\\\\
-3 < 2x - 1 <3\\\\
-3+1 < 2x - 1+1 <3+1\\\\
-2 < 2x <4\\\\
-2/2 < 2x/2 <4/2\\\\
-1 < x < 2\\\\[/tex]
If x is a nonnegative integer, then writing -1 < x < 2 is the exact same as writing the roster notation {0, 1}. This is the solution set. We ignore anything negative and it must be a whole number smaller than 2.
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If we were to plug in x = 0, then we'd get a true statement
[tex]|2x - 1| <3\\\\
|2*0 - 1| <3\\\\
|0 - 1| <3\\\\
|-1| <3\\\\
1 <3\\\\[/tex]
I'll let you try out x = 1. That should result in a true statement as well.
Trying out x = 2 or larger will result in a false statement.
