I’m preparing for a test and I keep getting stuck on number line reasoning with distances. For example: “Find all x that are closer to 3 than to −2.” I understand that distance on a number line is an absolute value, so I wrote |x − 3| < |x + 2|, but I’m not sure how to turn that into the correct shaded region without making sign mistakes. Should I split into cases around the key points (the two numbers and their midpoint), or is there a simpler approach that avoids piecewise analysis? I tried squaring both sides to remove the absolute values, but then I wasn’t sure about keeping the inequality direction and whether that introduces extra solutions. I also tried plotting both points and marking the midpoint visually, but I keep second-guessing which side to shade, especially when the wording changes to “at least as close” (do I include the midpoint or not, and why?). I also tried testing a few sample x-values, but that feels ad hoc and I’m not sure it’s reliable. If the statement flips (e.g., “closer to −2 than to 3”), does the region just switch sides, or do I need to re-check everything from scratch? Any help appreciated!
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3 Responses
Closer to a than to b means x is on the side of the midpoint m = (a + b)/2 toward a; since |·| ≥ 0 you can safely square |x − a| < |x − b| to get (x − a)^2 < (x − b)^2, which simplifies to x < m if a < b and x > m if a > b (include m for “at least as close”). Example: |x − 3| < |x + 2| ⇒ (x − 3)^2 < (x + 2)^2 ⇒ x > 1/2; for “at least as close,” x ≥ 1/2; reversed wording gives x < 1/2; see: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:absolute-value
No need to split cases-just take the midpoint m = (a + b)/2: points closer to a than b are the half-line on a’s side of m (exclude m for “closer,” include it for “at least as close”), so for 3 vs −2 the midpoint is 0.5 and “closer to 3” is x > 0.5 (or ≥ for “at least”), and flipping it to “closer to −2” gives x < 0.5 (or ≤). Hope this helps!
Think of the midpoint m = (a + b)/2 as the fence between team a and team b: “closer to a than to b” is everything on a’s side of that fence (|x−a| < |x−b| ⇔ (x−a)^2 < (x−b)^2 ⇒ x < m if a < b, x > m if a > b), and “at least as close” includes the fence (≤).
Example: a = 3, b = −2 gives m = 0.5, so “closer to 3 than to −2” is x > 0.5 (x ≥ 0.5 for “at least as close”); flipping to “closer to −2” gives x < 0.5.