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Question Number 110294 by bemath last updated on 28/Aug/20

∣2x+1∣−∣x−2∣ < 4 find the solution set

∣2x+1∣−∣x−2∣<4findthesolutionset

Answered by Rio Michael last updated on 28/Aug/20

∣2x + 1∣ = { ((2x + 1 , x ≥ −(1/2))),((−(2x + 1), x < −(1/2))) :} ∣x−2∣ = { ((x−2, x ≥ 2)),((−(x−2) , x < 2)) :} for x < −(1/2), −(2x + 1) −[−(x−2)] < 4 ⇒ −2x−1 + x−2 < 4 ⇒ −x −3 < 4 or x > −7 for −(1/2) ≤ x < 2, 2x + 1 + x−2 < 4 ⇒ 3x−1 < 4 or x < (5/3) for x ≥ 2, 2x + 1 − (x−2) <4 ⇒ 2x + 1 −x + 2 <4 x < 1 x > −7, x < (5/3) and x< 1 ⇒ −7 < x < (5/3)

∣2x+1∣={2x+1,x⩾−12−(2x+1),x<−12∣x−2∣={x−2,x⩾2−(x−2),x<2forx<−12,−(2x+1)−[−(x−2)]<4⇒−2x−1+x−2<4⇒−x−3<4orx>−7for−12⩽x<2,2x+1+x−2<4⇒3x−1<4orx<53forx⩾2,2x+1−(x−2)<4⇒2x+1−x+2<4x<1x>−7,x<53andx<1⇒−7<x<53

Commented by bemath last updated on 28/Aug/20

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