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Question Number 126180 by benjo_mathlover last updated on 18/Dec/20
solve ∣ ∣x−1∣ −2∣ = ∣ x−3 ∣
solve∣∣x−1∣−2∣=∣x−3∣
Answered by bobhans last updated on 18/Dec/20
(1) for x≥3 ⇒ ∣ x−3∣=x−3 (2) for x<3 ⇒∣1−x−2∣=3−x ∣−1−x∣ = 3−x (−1−x+3−x)(−1−x−3+x)=0 (2−2x)(−4)=0 ; x=1 solution {1}∪[3,∞)
(1)forx⩾3⇒∣x−3∣=x−3(2)forx<3⇒∣1−x−2∣=3−x∣−1−x∣=3−x(−1−x+3−x)(−1−x−3+x)=0(2−2x)(−4)=0;x=1solution{1}∪[3,∞)
Answered by mathmax by abdo last updated on 18/Dec/20
∣∣x−1∣−2∣=∣x−3∣ ⇔∣x−1∣−2=x−3(e_1 ) or ∣x−1∣−2=3−x(e_2 ) ∣x−1∣−2=x−3 ⇒∣x−1∣=x−1 ⇒x−1=x−1 or x−1=1−x ⇒ x=1 (e_2 )⇒∣x−1∣−2=3−x ⇒∣x−1∣=5−x (sox≤5) ⇒x−1=5−x or x−1=x−5 ⇒2x=6 or−1=−5(impo) ⇒x=3 so the set of solution is S={1,3}
∣∣x−1∣−2∣=∣x−3∣⇔∣x−1∣−2=x−3(e1)or∣x−1∣−2=3−x(e2)∣x−1∣−2=x−3⇒∣x−1∣=x−1⇒x−1=x−1orx−1=1−x⇒x=1(e2)⇒∣x−1∣−2=3−x⇒∣x−1∣=5−x(sox⩽5)⇒x−1=5−xorx−1=x−5⇒2x=6or−1=−5(impo)⇒x=3sothesetofsolutionisS={1,3}

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