silemus.

38Absolute value inequality

Solve |x + 3| < 5.

The solution is -8 < x < 2. Since |x+3| < 5 means x+3 is within 5 of zero, write -5 < x+3 < 5, then subtract 3 from every part to isolate x. Checking x=0: |0+3|=3 < 5 ✓.

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