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Question Number 175618 by daus last updated on 04/Sep/22

Commented by mr W last updated on 04/Sep/22

x≥0

x⩾0

Commented by mr W last updated on 05/Sep/22

Answered by greougoury555 last updated on 04/Sep/22

∣2x+1∣ ≥ ∣2x−1∣ 2x+1 ≥2x−1 ∪ 2x+1 ≤1−2x ∀xε R ∪ x≤ 0⇒ x≤0

∣2x+1∣⩾∣2x−1∣2x+1⩾2x−1∪2x+1⩽1−2x∀xϵR∪x⩽0⇒x⩽0

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