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?

a. −3<x<2-3 < x < 2,

b. −1<x<2-1 < x < 2,

c. −2<x<1-2 < x < 1,

d. −1.5<x<1-1.5 < x < 1,

e. NONO 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 alwaysalways true, we simply have −2<x<1-2 < x < 1, that is, we include the whole case.

Case 3: x≤−2.x \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.