Solve the Inequality for h |2h-3|>=9

$| 2 h - 3 | \geq 9$

Step 1

Write $| 2 h - 3 | \geq 9$ as a piecewise.

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

$2 h - 3 \geq 0$

Solve the inequality.

Add $3$ to both sides of the inequality.

$2 h \geq 3$

Divide each term in $2 h \geq 3$ by $2$ and simplify.

Divide each term in $2 h \geq 3$ by $2$ .

$\frac{2 h}{2} \geq \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 h}{2} \geq \frac{3}{2}$

Divide $h$ by $1$ .

$h \geq \frac{3}{2}$

In the piece where $2 h - 3$ is non-negative, remove the absolute value.

$2 h - 3 \geq 9$

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

$2 h - 3 < 0$

Solve the inequality.

Add $3$ to both sides of the inequality.

$2 h < 3$

Divide each term in $2 h < 3$ by $2$ and simplify.

Divide each term in $2 h < 3$ by $2$ .

$\frac{2 h}{2} < \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 h}{2} < \frac{3}{2}$

Divide $h$ by $1$ .

$h < \frac{3}{2}$

In the piece where $2 h - 3$ is negative, remove the absolute value and multiply by $- 1$ .

$- (2 h - 3) \geq 9$

Write as a piecewise.

${\begin{cases} 2 h - 3 \geq 9 & h \geq \frac{3}{2} \\ - (2 h - 3) \geq 9 & h < \frac{3}{2} \end{cases}$

Simplify $- (2 h - 3) \geq 9$ .

Apply the distributive property.

${\begin{cases} 2 h - 3 \geq 9 & h \geq \frac{3}{2} \\ - (2 h) - - 3 \geq 9 & h < \frac{3}{2} \end{cases}$

Multiply $2$ by $- 1$ .

${\begin{cases} 2 h - 3 \geq 9 & h \geq \frac{3}{2} \\ - 2 h - - 3 \geq 9 & h < \frac{3}{2} \end{cases}$

Multiply $- 1$ by $- 3$ .

${\begin{cases} 2 h - 3 \geq 9 & h \geq \frac{3}{2} \\ - 2 h + 3 \geq 9 & h < \frac{3}{2} \end{cases}$

Step 2

Solve $2 h - 3 \geq 9$ for $h$ .

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

Add $3$ to both sides of the inequality.

$2 h \geq 9 + 3$

Add $9$ and $3$ .

$2 h \geq 12$

Divide each term in $2 h \geq 12$ by $2$ and simplify.

Divide each term in $2 h \geq 12$ by $2$ .

$\frac{2 h}{2} \geq \frac{12}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 h}{2} \geq \frac{12}{2}$

Divide $h$ by $1$ .

$h \geq \frac{12}{2}$

Simplify the right side.

Divide $12$ by $2$ .

$h \geq 6$

Step 3

Solve $- 2 h + 3 \geq 9$ for $h$ .

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

Subtract $3$ from both sides of the inequality.

$- 2 h \geq 9 - 3$

Subtract $3$ from $9$ .

$- 2 h \geq 6$

Divide each term in $- 2 h \geq 6$ by $- 2$ and simplify.

Divide each term in $- 2 h \geq 6$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 h}{- 2} \leq \frac{6}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 h}{- 2} \leq \frac{6}{- 2}$

Divide $h$ by $1$ .

$h \leq \frac{6}{- 2}$

Simplify the right side.

Divide $6$ by $- 2$ .

$h \leq - 3$

Step 4

Find the union of the solutions.

$h \leq - 3$ or $h \geq 6$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$h \leq - 3 or h \geq 6$

Interval Notation:

$(- \infty, - 3] \cup [6, \infty)$

Step 6

Solve the Inequality for h |2h-3|>=9

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