1. **State the problem:** Find the range of values for which $x+5 < |2x + 1|$. 2. **Understand the absolute value:** The inequality involves an absolute value, so we consider two cases: - Case 1: $2x + 1 \geq 0$ so $|2x + 1| = 2x + 1$ - Case 2: $2x + 1 < 0$ so $|2x + 1| = -(2x + 1) = -2x - 1$ 3. **Case 1: $2x + 1 \geq 0$** This means $x \geq -\frac{1}{2}$. The inequality becomes: $$x + 5 < 2x + 1$$ Subtract $x$ from both sides: $$5 < x + 1$$ Subtract 1 from both sides: $$4 < x$$ So for Case 1, the solution is: $$x > 4$$ But remember $x \geq -\frac{1}{2}$, so the combined condition is: $$x > 4$$ 4. **Case 2: $2x + 1 < 0$** This means $x < -\frac{1}{2}$. The inequality becomes: $$x + 5 < -2x - 1$$ Add $2x$ to both sides: $$3x + 5 < -1$$ Subtract 5 from both sides: $$3x < -6$$ Divide both sides by 3: $$x < -2$$ So for Case 2, the solution is: $$x < -2$$ Since $x < -\frac{1}{2}$ is already true here, the combined condition is: $$x < -2$$ 5. **Combine both cases:** The solution to the inequality is: $$x < -2 \quad \text{or} \quad x > 4$$ **Final answer:** The range of values for which $x+5 < |2x + 1|$ is $$(-\infty, -2) \cup (4, \infty)$$