Solve: $$ |2 x+1| \geq 5 $$ Glve your answer as an interval or union of intervals. Only use integers or fractions, no decimal

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$$ (-\infty, -3] \cup [2, \infty) $$ To solve the inequality $$ |2x + 1| \geq 5 $$, we need to consider the definition of absolute value. The expression $$ |A| \geq B $$ implies two cases: 1. $$ A \geq B $$ 2. $$ A \leq -B $$ In our case, we have: 1. $$ 2x + 1 \geq 5 $$ 2. $$ 2x + 1 \leq -5 $$ Let's solve each case separately. **Case 1: $$ 2x + 1 \geq 5 $$** Subtract 1 from both sides: $$ 2x \geq 4 $$ Now, divide by 2: $$ x \geq 2 $$ **Case 2: $$ 2x + 1 \leq -5 $$** Subtract 1 from both sides: $$ 2x \leq -6 $$ Now, divide by 2: $$ x \leq -3 $$ Now we combine the results from both cases. The solution to the inequality $$ |2x + 1| \geq 5 $$ is: $$ x \leq -3 \quad \text{or} \quad x \geq 2 $$ In interval notation, this is expressed as: $$ (-\infty, -3] \cup [2, \infty) $$ Thus, the final answer is: $$ \boxed{(-\infty, -3] \cup [2, \infty)} $$

Solve: \( |2 x+1| \geq 5 \) Glve your answer as an interval or union of intervals. Only use integers or fractions, no decimal

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Solve: \( |2 x+1| \geq 5 \) Glve your answer as an interval or union of intervals. Only use integers or fractions, no decimal

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