Solve for x |7-2x|>9

$| 7 - 2 x | > 9$

Step 1

Write $| 7 - 2 x | > 9$ as a piecewise.

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

$7 - 2 x \geq 0$

Solve the inequality.

Subtract $7$ from both sides of the inequality.

$- 2 x \geq - 7$

Divide each term in $- 2 x \geq - 7$ by $- 2$ and simplify.

Divide each term in $- 2 x \geq - 7$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} \leq \frac{- 7}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} \leq \frac{- 7}{- 2}$

Divide $x$ by $1$ .

$x \leq \frac{- 7}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{7}{2}$

In the piece where $7 - 2 x$ is non-negative, remove the absolute value.

$7 - 2 x > 9$

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

$7 - 2 x < 0$

Solve the inequality.

Subtract $7$ from both sides of the inequality.

$- 2 x < - 7$

Divide each term in $- 2 x < - 7$ by $- 2$ and simplify.

Divide each term in $- 2 x < - 7$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} > \frac{- 7}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} > \frac{- 7}{- 2}$

Divide $x$ by $1$ .

$x > \frac{- 7}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x > \frac{7}{2}$

In the piece where $7 - 2 x$ is negative, remove the absolute value and multiply by $- 1$ .

$- (7 - 2 x) > 9$

Write as a piecewise.

${\begin{cases} 7 - 2 x > 9 & x \leq \frac{7}{2} \\ - (7 - 2 x) > 9 & x > \frac{7}{2} \end{cases}$

Simplify $- (7 - 2 x) > 9$ .

Apply the distributive property.

${\begin{cases} 7 - 2 x > 9 & x \leq \frac{7}{2} \\ - 1 \cdot 7 - (- 2 x) > 9 & x > \frac{7}{2} \end{cases}$

Multiply $- 1$ by $7$ .

${\begin{cases} 7 - 2 x > 9 & x \leq \frac{7}{2} \\ - 7 - (- 2 x) > 9 & x > \frac{7}{2} \end{cases}$

Multiply $- 2$ by $- 1$ .

${\begin{cases} 7 - 2 x > 9 & x \leq \frac{7}{2} \\ - 7 + 2 x > 9 & x > \frac{7}{2} \end{cases}$

Step 2

Solve $7 - 2 x > 9$ for $x$ .

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

Subtract $7$ from both sides of the inequality.

$- 2 x > 9 - 7$

Subtract $7$ from $9$ .

$- 2 x > 2$

Divide each term in $- 2 x > 2$ by $- 2$ and simplify.

Divide each term in $- 2 x > 2$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} < \frac{2}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} < \frac{2}{- 2}$

Divide $x$ by $1$ .

$x < \frac{2}{- 2}$

Simplify the right side.

Divide $2$ by $- 2$ .

$x < - 1$

Step 3

Solve $- 7 + 2 x > 9$ for $x$ .

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

Add $7$ to both sides of the inequality.

$2 x > 9 + 7$

Add $9$ and $7$ .

$2 x > 16$

Divide each term in $2 x > 16$ by $2$ and simplify.

Divide each term in $2 x > 16$ by $2$ .

$\frac{2 x}{2} > \frac{16}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{16}{2}$

Divide $x$ by $1$ .

$x > \frac{16}{2}$

Simplify the right side.

Divide $16$ by $2$ .

$x > 8$

Step 4

Find the union of the solutions.

$x < - 1$ or $x > 8$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or x > 8$

Interval Notation:

$(- \infty, - 1) \cup (8, \infty)$

Step 6

Solve for x |7-2x|>9

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