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Solve: |x - 3| < 8
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X is any number greater than -5 but less than 11 |x-3| < 8 (rewrite the equation to get read of the absolution) -8 < x-3 < 8 (add three to both sides of the inequality to get x) -5 < x < 11 If we substitute any number between -5 and 11 we see this is true |x-3| < 8 if x = -4 or |x-3| < 8 if x = 10 |-4-3| < 8 ― |10-3| < 8 |-7| < 8 ― |7| < 8 7 < 8 (true) ― 7<8 (true)
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