1. Problem: Solve the inequality $|2x + 3| \geq 3$. 2. Recall the definition of absolute value inequality: $|A| \geq B$ means $A \geq B$ or $A \leq -B$ for $B \geq 0$. 3. Apply this to $|2x + 3| \geq 3$: $$2x + 3 \geq 3 \quad \text{or} \quad 2x + 3 \leq -3$$ 4. Solve each inequality separately: - For $2x + 3 \geq 3$: $$2x + 3 \geq 3$$ $$2x \geq 3 - 3$$ $$2x \geq 0$$ $$\cancel{2}x \geq \cancel{0}$$ $$x \geq 0$$ - For $2x + 3 \leq -3$: $$2x + 3 \leq -3$$ $$2x \leq -3 - 3$$ $$2x \leq -6$$ $$\cancel{2}x \leq \cancel{-6}$$ $$x \leq -3$$ 5. Combine the solution sets: $$x \leq -3 \quad \text{or} \quad x \geq 0$$ 6. Final answer: The solution to the inequality $|2x + 3| \geq 3$ is $$(-\infty, -3] \cup [0, \infty)$$ This means all $x$ less than or equal to $-3$ and all $x$ greater than or equal to $0$. --- "slug": "abs inequality", "subject": "algebra", "desmos": { "latex": "y=|2x+3|-3", "features": { "intercepts": true, "extrema": false } }, "q_count": 1