Question

The solution set of the inequality $$ |3x| \geq |6 - 3x| $$ is: 

• A. (−∞, 1]
• B. [1, ∞)
• C. (−∞, 1) ∪ (1,∞)
• D. (−∞, −1) ∪ (−1,∞)

Hint:

When solving inequalities involving absolute values, it’s important to break them into different cases based on the possible signs of the expressions inside the absolute values. Each case leads to a different inequality that you can solve. If you end up with a contradiction (like \( 0 \leq -6 \)), that case provides no valid solutions. Once all cases are considered, the union of their solutions will give you the final answer.

Answer 1

The Correct Option is B

The inequality \( |3x| \geq |6 - 3x| \) involves absolute values, so split into cases:

Case 1: \( 3x \geq 6 - 3x \):

\( 3x + 3x \geq 6 \implies 6x \geq 6 \implies x \geq 1 \).

Case 2: \( 3x \leq -(6 - 3x) \):

\( 3x \leq -6 + 3x \implies 0 \leq -6 \),

which is not possible.

Thus, the solution is \( x \geq 1 \), or \([1, \infty)\).

Answer 2

The inequality \( |3x| \geq |6 - 3x| \) involves absolute values, so we need to split it into cases based on the definitions of absolute value:

Step 1: Case 1 - \( 3x \geq 6 - 3x \):

In this case, we assume that the expressions inside the absolute values are positive. Solving this inequality: \[ 3x + 3x \geq 6 \implies 6x \geq 6 \implies x \geq 1. \] Therefore, the solution for this case is \( x \geq 1 \).

Step 2: Case 2 - \( 3x \leq -(6 - 3x) \):

In this case, we assume that the expression inside the absolute value on the right side is negative. Solving this inequality: \[ 3x \leq -6 + 3x \implies 0 \leq -6, \] which is a contradiction (since 0 is not less than or equal to -6). Thus, this case has no solution.

Step 3: Conclusion:

From Case 1, we get \( x \geq 1 \), and Case 2 does not contribute any solutions. Therefore, the solution to the inequality is: \[ x \geq 1 \quad \text{or} \quad [1, \infty). \]