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3 Responses
Great question! Since |2x−1| is always nonnegative, any solution must satisfy x ≥ 0-but you don’t have to impose that at the very start as long as you check your answers at the end. The cleanest way is to split by the sign of 2x−1: if 2x−1 ≥ 0 (so x ≥ 1/2), then |2x−1| = 2x−1 and the equation 2x−1 = x gives x = 1, which fits x ≥ 1/2. If 2x−1 < 0 (so x < 1/2), then |2x−1| = 1−2x and 1−2x = x gives x = 1/3, which fits x < 1/2. Both are ≥ 0, so both work. Your split 2x−1 = x or 2x−1 = −x is equivalent-you just need to toss any negative x that might appear. Final solutions: x = 1/3 and x = 1. Hope this helps!
Great question! Think of |2x−1| as a distance on a number line-so if it equals x, then x itself can’t be negative (you can’t walk −5 meters!), which means any solution must satisfy x ≥ 0. If you use the “|A| = x means A = x or A = −x” shortcut, then yes, you should include x ≥ 0 to make it logically correct. But you don’t have to front-load that if you split by the sign of the inside instead: if 2x−1 ≥ 0 (so x ≥ 1/2), then 2x−1 = x gives x = 1, which fits; if 2x−1 < 0 (so x < 1/2), then 1−2x = x gives x = 1/3, which also fits and is nonnegative. So the solutions are x = 1/3 and x = 1. You’re not overthinking-your instinct about x ≥ 0 is spot on-but checking by cases on 2x−1 handles it cleanly. Hope this helps!
You don’t need to assume x ≥ 0 first. Solve 2x−1 = x and 2x−1 = −x, then discard any negative x (since |·| ≥ 0); this gives x = 1 and x = 1/3, though I’m not 100% sure that split is the cleanest way.