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Question Number 126180 by benjo_mathlover last updated on 18/Dec/20
solve∣∣x−1∣−2∣=∣x−3∣
Answered by bobhans last updated on 18/Dec/20
(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(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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