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Question Number 126180 by benjo_mathlover last updated on 18/Dec/20

solve ∣ ∣x−1∣ −2∣ = ∣ x−3 ∣

solvex12=x3

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)forx3x3∣=x3(2)forx<3⇒∣1x2∣=3x1x=3x(1x+3x)(1x3+x)=0(22x)(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}

∣∣x12∣=∣x3⇔∣x12=x3(e1)orx12=3x(e2)x12=x3⇒∣x1∣=x1x1=x1orx1=1xx=1(e2)⇒∣x12=3x⇒∣x1∣=5x(sox5)x1=5xorx1=x52x=6or1=5(impo)x=3sothesetofsolutionisS={1,3}

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