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| Opening Absolute Value using |x| = x when x≥0 and |x| = -x when x ≤ 0 | |
-x ≥ 0 => |x| = x => x = 5x - 16 => 4x = 16 => x = \(\frac{16}{4}\) = 4 And 4 > 0 => x = 4 is a solution | -x ≤ 0 => |x| = -x => -x = 5x - 16 => 6x = 16 => x = \(\frac{16}{6}\) = \(\frac{8}{3}\) But \(\frac{8}{3}\) is NOT < 0 => x = \(\frac{8}{3}\) is NOT a solution |
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