Answer: Choice A
Open holes at 0 and 2; shading in between.
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Work Shown:
|3x - 3| < 3
- 3 < 3x - 3 < 3
- 3+3 < 3x - 3+3 < 3+3
0 < 3x < 6
0/3 < 3x/3 < 6/3
0 < x < 2
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Step-by-step explanation:
The rule to use is when |x| < k, then -k < x < k where k is positive.
After that rule is used, I added 3 to all sides and then divided all sides by 3 to isolate x.
The graph of 0 < x < 2 will have open holes at 0 and 2; then shade in between those open holes.
This describes "x is anything between 0 and 2 excluding both endpoints".
Therefore, we'll go for answer choice A
Choice C is close, but the endpoints should be open holes to indicate "exclude these endpoints". This means x = 0 and x = 2 are not part of the shaded solution set. If "or equal to" was part of the inequality signs, then choice C would be the answer.