Question
Solve for x: |2x − 5| + |x + 1| ≤
10.Solution
ATQ,
Critical points at x = 2.5 and x = −1. Case 1: x ≥ 2.5 ⇒ (2x − 5) + (x + 1) ≤ 10 ⇒ 3x − 4 ≤ 10 ⇒ x ≤ 14/3. So 2.5 ≤ x ≤ 14/3. Case 2: −1 ≤ x ≤ 2.5 ⇒ (5 − 2x) + (x + 1) ≤ 10 ⇒ 6 − x ≤ 10 ⇒ −x ≤ 4 ⇒ x ≥ −4. In this band, it’s all valid: −1 ≤ x ≤ 2.5. Case 3: x ≤ −1 ⇒ (5 − 2x) + (−x − 1) ≤ 10 ⇒ 4 − 3x ≤ 10 ⇒ −3x ≤ 6 ⇒ x ≥ −2. So in this band: −2 ≤ x ≤ −1. Combine: −2 ≤ x ≤ 14/3. Answer: x ∈ [−2, 14/3].
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