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2012 Math Kangaroo

Problem 22

Problem 22 · 2012 Math Kangaroo Stretch
Algebra & Patterns casework

The solution set of the inequality \(|x| + |x-3| > 3\) is

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Answer: A — \(\left]-\infty, 0\right[ \cup \left]3, +\infty\right[\)
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Hint 1 of 2
Read \(|x| + |x-3|\) as the distance from \(x\) to \(0\) plus the distance from \(x\) to \(3\).
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Hint 2 of 2
That total bottoms out at \(3\) for \(x\) between 0 and 3; ask when it strictly exceeds 3.
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Approach: read |x|+|x−3| as a sum of distances
  1. \(|x| + |x-3|\) is the distance from \(x\) to \(0\) plus the distance from \(x\) to \(3\).
  2. For any \(x\) between 0 and 3 that sum equals exactly \(3\); outside the interval it grows larger.
  3. So the sum exceeds 3 exactly when \(x < 0\) or \(x > 3\), i.e. \(\left]-\infty,0\right[ \cup \left]3,+\infty\right[\), choice A.
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