Question

which of the following is the solution to $3|x - 1| \geq 12$? \
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\\(\bigcirc\\) a. $x \leq -3$ and $x \geq 5$ \
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\\(\bigcirc\\) b. $x \geq -3$ or $x \geq 5$ \
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\\(\bigcirc\\) c. $x \geq 5$ \
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\\(\bigcirc\\) d. $x \leq -3$ or $x \geq 5$

Explanation:

Step1: Divide both sides by 3

To solve the inequality \( 3|x - 1| \geq 12 \), we first divide both sides by 3 to isolate the absolute value expression.

$$ \frac{3|x - 1|}{3} \geq \frac{12}{3} $$

Simplifying both sides, we get:

$$ |x - 1| \geq 4 $$

Step2: Solve the absolute value inequality

The absolute value inequality \( |x - 1| \geq 4 \) means that \( x - 1 \) is either less than or equal to -4 or greater than or equal to 4. This gives us two separate inequalities:

1. \( x - 1 \leq -4 \)
2. \( x - 1 \geq 4 \)

Solving the first inequality (\( x - 1 \leq -4 \)):

Add 1 to both sides:

$$ x - 1 + 1 \leq -4 + 1 $$

Simplifying, we get:

$$ x \leq -3 $$

Solving the second inequality (\( x - 1 \geq 4 \)):

Add 1 to both sides:

$$ x - 1 + 1 \geq 4 + 1 $$

Simplifying, we get:

$$ x \geq 5 $$

Combining these two solutions, we have \( x \leq -3 \) or \( x \geq 5 \).

Answer:

D. \( x \leq -3 \) or \( x \geq 5 \)