Solve for x 3|x-1|>=12

$3 | x - 1 | \geq 12$

Step 1

Write $3 | x - 1 | \geq 12$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x - 1 \geq 0$

Add $1$ to both sides of the inequality.

$x \geq 1$

In the piece where $x - 1$ is non-negative, remove the absolute value.

$3 (x - 1) \geq 12$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x - 1 < 0$

Add $1$ to both sides of the inequality.

$x < 1$

In the piece where $x - 1$ is negative, remove the absolute value and multiply by $- 1$ .

$3 (- (x - 1)) \geq 12$

Write as a piecewise.

${\begin{cases} 3 (x - 1) \geq 12 & x \geq 1 \\ 3 (- (x - 1)) \geq 12 & x < 1 \end{cases}$

Simplify $3 (x - 1) \geq 12$ .

Apply the distributive property.

${\begin{cases} 3 x + 3 \cdot - 1 \geq 12 & x \geq 1 \\ 3 (- (x - 1)) \geq 12 & x < 1 \end{cases}$

Multiply $3$ by $- 1$ .

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ 3 (- (x - 1)) \geq 12 & x < 1 \end{cases}$

Simplify $3 (- (x - 1)) \geq 12$ .

Apply the distributive property.

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ 3 (- x - - 1) \geq 12 & x < 1 \end{cases}$

Multiply $- 1$ by $- 1$ .

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ 3 (- x + 1) \geq 12 & x < 1 \end{cases}$

Apply the distributive property.

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ 3 (- x) + 3 \cdot 1 \geq 12 & x < 1 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ - 3 x + 3 \cdot 1 \geq 12 & x < 1 \end{cases}$

Multiply $3$ by $1$ .

${\begin{cases} 3 x - 3 \geq 12 & x \geq 1 \\ - 3 x + 3 \geq 12 & x < 1 \end{cases}$

Step 2

Solve $3 x - 3 \geq 12$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $3$ to both sides of the inequality.

$3 x \geq 12 + 3$

Add $12$ and $3$ .

$3 x \geq 15$

Divide each term in $3 x \geq 15$ by $3$ and simplify.

Divide each term in $3 x \geq 15$ by $3$ .

$\frac{3 x}{3} \geq \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} \geq \frac{15}{3}$

Divide $x$ by $1$ .

$x \geq \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$x \geq 5$

Step 3

Solve $- 3 x + 3 \geq 12$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $3$ from both sides of the inequality.

$- 3 x \geq 12 - 3$

Subtract $3$ from $12$ .

$- 3 x \geq 9$

Divide each term in $- 3 x \geq 9$ by $- 3$ and simplify.

Divide each term in $- 3 x \geq 9$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} \leq \frac{9}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} \leq \frac{9}{- 3}$

Divide $x$ by $1$ .

$x \leq \frac{9}{- 3}$

Simplify the right side.

Divide $9$ by $- 3$ .

$x \leq - 3$

Step 4

Find the union of the solutions.

$x \leq - 3$ or $x \geq 5$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or x \geq 5$

Interval Notation:

$(- \infty, - 3] \cup [5, \infty)$

Step 6

Solve for x 3|x-1|>=12

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