1973 AMC 12 Problem 22

Attempt Problem 22 of the 1973 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1973 AMC 12 solutions, or check the answer key.

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22.

The set of all real solutions of the inequality ∣x−1∣+∣x+2∣<5 |x-1|+|x+2|\lt5 is
{x:−3<x<2}\{x:-3\lt x\lt2\}
{x:−1<x<2}\{x:-1\lt x\lt2\}
{x:−2<x<1}\{x:-2\lt x\lt1\}
{x:−32<x<72}\left\{x:-\dfrac32\lt x\lt\dfrac72\right\}
∅\varnothing (empty)
Answer: A
Concepts:absolute valueinequality
Small hint:
Interpret the two absolute values as the distances from xx to 11 and to −2-2
Big hint:
Between −2-2 and 11 the sum is constant; outside that interval it increases by twice the distance from the nearer endpoint
Solution:
For −2≤x≤1,-2\le x\le1, the sum of the distances from xx to −2-2 and 11 is 3.3. If xx lies a distance uu outside this interval, the sum is 3+2u.3+2u. Thus 3+2u<5⟺u<1. 3+2u\lt5 \quad\Longleftrightarrow\quad u\lt1. Extending the interval [−2,1][-2,1] by 11 at each end gives −3<x<2. -3\lt x\lt2.
Therefore, the correct answer is A.
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