2211.11 – Absolutely Unequal

Which of the following gives the set of real solutions to the inequality: ∣x−1∣+∣x+2∣<5? \vert x-1\vert + \vert x+2\vert < 5?

  1. −3<x<2-3 < x < 2

  2. −1<x<2-1 < x < 2

  3. −2<x<1-2 < x < 1

  4. −1.5<x<1-1.5 < x < 1

  5. No real solution.

Solution

There are three cases.

Case 1: x≥1x \geq 1. Then:

∣x−1∣+∣x+2∣<5x−1+x+2<5,2x+1<5,2x<4,x<2.\begin{aligned}\vert x-1\vert + \vert x+2\vert &< 5 \\x-1+x+2 &< 5, \\2x+1 &< 5, \\2x&<4,\\x&<2.\end{aligned}

In summary: 1≤x<21\leq x<2.

Case 2: −2<x<1-2 < x < 1. Then:

∣x−1∣+∣x+2∣<51−x+x+2<5,3<5.\begin{aligned}\vert x-1\vert + \vert x+2\vert &< 5 \\1-x+x+2 &< 5, \\3 &< 5.\end{aligned}

Since this is always true, we simply have −2<x<1-2 < x < 1, that is, we include the whole case.

Case 3: x≤−2x \leq -2. Then:

∣x−1∣+∣x+2∣<51−x−x−2<5,−2x−1<5,−2x<6,x>−3.\begin{aligned}\vert x-1\vert + \vert x+2\vert &< 5 \\1-x-x-2 &< 5, \\-2x-1 &< 5, \\-2x &< 6, \\x &> -3.\end{aligned}

In summary, −3<x≤−2-3 < x \leq -2.

Putting the three cases together: −3<x<2-3 < x < 2. The answer is a.