Solve the Inequality for x 3|x-5|>9

$3 | x - 5 | > 9$

Write $3 | x - 5 | > 9$ as a piecewise.

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

$x - 5 \geq 0$

Add $5$ to both sides of the inequality.

$x \geq 5$

In the piece where $x - 5$ is non-negative, remove the absolute value.

$3 (x - 5) > 9$

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

$x - 5 < 0$

Add $5$ to both sides of the inequality.

$x < 5$

In the piece where $x - 5$ is negative, remove the absolute value and multiply by $- 1$ .

$3 (- (x - 5)) > 9$

Write as a piecewise.

${\begin{cases} 3 (x - 5) > 9 & x \geq 5 \\ 3 (- (x - 5)) > 9 & x < 5 \end{cases}$

Simplify $3 (x - 5) > 9$ .

Apply the distributive property.

${\begin{cases} 3 x + 3 \cdot - 5 > 9 & x \geq 5 \\ 3 (- (x - 5)) > 9 & x < 5 \end{cases}$

Multiply $3$ by $- 5$ .

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ 3 (- (x - 5)) > 9 & x < 5 \end{cases}$

Simplify $3 (- (x - 5)) > 9$ .

Apply the distributive property.

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ 3 (- x - - 5) > 9 & x < 5 \end{cases}$

Multiply $- 1$ by $- 5$ .

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ 3 (- x + 5) > 9 & x < 5 \end{cases}$

Apply the distributive property.

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ 3 (- x) + 3 \cdot 5 > 9 & x < 5 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ - 3 x + 3 \cdot 5 > 9 & x < 5 \end{cases}$

Multiply $3$ by $5$ .

${\begin{cases} 3 x - 15 > 9 & x \geq 5 \\ - 3 x + 15 > 9 & x < 5 \end{cases}$

Solve $3 x - 15 > 9$ for $x$ .

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

Add $15$ to both sides of the inequality.

$3 x > 9 + 15$

Add $9$ and $15$ .

$3 x > 24$

Divide each term in $3 x > 24$ by $3$ and simplify.

Divide each term in $3 x > 24$ by $3$ .

$\frac{3 x}{3} > \frac{24}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} > \frac{24}{3}$

Divide $x$ by $1$ .

$x > \frac{24}{3}$

Simplify the right side.

Divide $24$ by $3$ .

$x > 8$

Solve $- 3 x + 15 > 9$ for $x$ .

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

Subtract $15$ from both sides of the inequality.

$- 3 x > 9 - 15$

Subtract $15$ from $9$ .

$- 3 x > - 6$

Divide each term in $- 3 x > - 6$ by $- 3$ and simplify.

Divide each term in $- 3 x > - 6$ by $- 3$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 3 x}{- 3} < \frac{- 6}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} < \frac{- 6}{- 3}$

Divide $x$ by $1$ .

$x < \frac{- 6}{- 3}$

Simplify the right side.

Divide $- 6$ by $- 3$ .

$x < 2$

Find the union of the solutions.

$x < 2$ or $x > 8$

The result can be shown in multiple forms.

Inequality Form:

$x < 2 or x > 8$

Interval Notation:

$(- \infty, 2) \cup (8, \infty)$

Solve the Inequality for x 3|x-5|>9

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