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| |x+10|=2x+8 , so we will have two cases | |
| -Case 1: x + 10 ≥ 0, => x ≥ -10 => |x + 10| = x + 10 => x + 10 = 2x + 8 => x = 2 But the condition was x ≥ -10 and 2 ≥ -10 => x = 2 is a SOLUTION | -Case 2: x + 10 ≤ 0 => x ≤ -10 |x + 10| = -(x + 10) = -x - 10 => -x - 10 = 2x + 8 => 3x = -18 => x = \(\frac{-18}{3}\) = -6 But the condition was x ≤ -10 and -6 IS NOT ≤ -10 => x = -6 is NOT a SOLUTION |


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