Convert to Interval Notation 5/(2x-9)>1

$\frac{5}{2 x - 9} > 1$

Move $1$ to the left side of the equation by subtracting it from both sides.

$\frac{5}{2 x - 9} - 1 > 0$

Simplify $\frac{5}{2 x - 9} - 1$ .

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To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2 x - 9}{2 x - 9}$ .

$\frac{5}{2 x - 9} - 1 \cdot \frac{2 x - 9}{2 x - 9} > 0$

Simplify terms.

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Combine $- 1$ and $\frac{2 x - 9}{2 x - 9}$ .

$\frac{5}{2 x - 9} + \frac{- (2 x - 9)}{2 x - 9} > 0$

Combine the numerators over the common denominator.

$\frac{5 - (2 x - 9)}{2 x - 9} > 0$

Simplify the numerator.

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Apply the distributive property.

$\frac{5 - (2 x) - - 9}{2 x - 9} > 0$

Multiply $2$ by $- 1$ .

$\frac{5 - 2 x - - 9}{2 x - 9} > 0$

Multiply $- 1$ by $- 9$ .

$\frac{5 - 2 x + 9}{2 x - 9} > 0$

Add $5$ and $9$ .

$\frac{- 2 x + 14}{2 x - 9} > 0$

Factor $2$ out of $- 2 x + 14$ .

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Factor $2$ out of $- 2 x$ .

$\frac{2 (- x) + 14}{2 x - 9} > 0$

Factor $2$ out of $14$ .

$\frac{2 (- x) + 2 (7)}{2 x - 9} > 0$

Factor $2$ out of $2 (- x) + 2 (7)$ .

$\frac{2 (- x + 7)}{2 x - 9} > 0$

Simplify with factoring out.

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Factor $- 1$ out of $- x$ .

$\frac{2 (- (x) + 7)}{2 x - 9} > 0$

Rewrite $7$ as $- 1 (- 7)$ .

$\frac{2 (- (x) - 1 (- 7))}{2 x - 9} > 0$

Factor $- 1$ out of $- (x) - 1 (- 7)$ .

$\frac{2 (- (x - 7))}{2 x - 9} > 0$

Simplify the expression.

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Rewrite $- (x - 7)$ as $- 1 (x - 7)$ .

$\frac{2 (- 1 (x - 7))}{2 x - 9} > 0$

Move the negative in front of the fraction.

$- \frac{2 (x - 7)}{2 x - 9} > 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x - 7 = 0$

$2 x - 9 = 0$

Add $7$ to both sides of the equation.

$x = 7$

Add $9$ to both sides of the equation.

$2 x = 9$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 9$ by $2$ .

$\frac{2 x}{2} = \frac{9}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{9}{2}$

Divide $x$ by $1$ .

$x = \frac{9}{2}$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 7$

$x = \frac{9}{2}$

Consolidate the solutions.

$x = 7, \frac{9}{2}$

Find the domain of $- \frac{2 (x - 7)}{2 x - 9}$ .

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Set the denominator in $\frac{2 (x - 7)}{2 x - 9}$ equal to $0$ to find where the expression is undefined.

$2 x - 9 = 0$

Solve for $x$ .

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Add $9$ to both sides of the equation.

$2 x = 9$

Divide each term by $2$ and simplify.

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Divide each term in $2 x = 9$ by $2$ .

$\frac{2 x}{2} = \frac{9}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{9}{2}$

Divide $x$ by $1$ .

$x = \frac{9}{2}$

The domain is all values of $x$ that make the expression defined.

$(- \infty, \frac{9}{2}) \cup (\frac{9}{2}, \infty)$

Use each root to create test intervals.

$x < \frac{9}{2}$

$\frac{9}{2} < x < 7$

$x > 7$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Test a value on the interval $x < \frac{9}{2}$ to see if it makes the inequality true.

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Choose a value on the interval $x < \frac{9}{2}$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{5}{2 (0) - 9} > 1$

The left side $- 0.\overline{5}$ is not greater than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $\frac{9}{2} < x < 7$ to see if it makes the inequality true.

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Choose a value on the interval $\frac{9}{2} < x < 7$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{5}{2 (6) - 9} > 1$

The left side $1.\overline{6}$ is greater than the right side $1$ , which means that the given statement is always true.

True

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$\frac{5}{2 (10) - 9} > 1$

The left side $0.\overline{45}$ is not greater than the right side $1$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$x < \frac{9}{2}$ False

$\frac{9}{2} < x < 7$ True

$x > 7$ False

$x < \frac{9}{2}$ False

$\frac{9}{2} < x < 7$ True

$x > 7$ False

The solution consists of all of the true intervals.

$\frac{9}{2} < x < 7$

Convert the inequality to interval notation.

$(\frac{9}{2}, 7)$

Convert to Interval Notation 5/(2x-9)>1

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