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Solve $|x-3|+|x-2|=1$ |
1 ≤ x ≥ 3 2 ≤ x ≤ 3 2 ≤ x ≤ 5 2 ≥ x ≥ 4 |
2 ≤ x ≤ 3 |
$|x-3|+|x-2|=1$ $⇒|x-3|+|x-2|=(3-x)+(x-2)$ ⇒ x - 3 ≤ 0 and x - 2 ≥ 0 ⇒ x ≤ 3 and x ≥ 2 ⇒ 2 ≤ x ≤ 3 |
