Solve ∣2x−3∣≤5|2x - 3| \le 5∣2x−3∣≤5.
Using property ∣X∣≤a ⟹ −a≤X≤a|X| \le a \implies -a \le X \le a∣X∣≤a⟹−a≤X≤a: −5≤2x−3≤5-5 \le 2x - 3 \le 5−5≤2x−3≤5 Add 3 to all parts: −2≤2x≤8-2 \le 2x \le 8−2≤2x≤8 Divide by 2: −1≤x≤4-1 \le x \le 4−1≤x≤4 Interval: [−1,4][-1, 4][−1,4].
Solve ∣x−1∣≥3|x - 1| \ge 3∣x−1∣≥3.
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