Solve the Inequality for x (x^2-1)/(3x-6)>0

$\frac{x^{2} - 1}{3 x - 6} > 0$

Step 1

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

$x^{2} - 1 = 0$

$3 x - 6 = 0$

Step 2

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{1}$

Step 4

Any root of $1$ is $1$ .

$x = \pm 1$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Step 6

Add $6$ to both sides of the equation.

$3 x = 6$

Step 7

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

Divide each term in $3 x = 6$ by $3$ .

$\frac{3 x}{3} = \frac{6}{3}$

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 $3$ .

$x = 2$

Step 8

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

$x = 1, - 1$

$x = 2$

Step 9

Consolidate the solutions.

$x = 1, - 1, 2$

Step 10

Find the domain of $\frac{x^{2} - 1}{3 x - 6}$ .

Set the denominator in $\frac{x^{2} - 1}{3 x - 6}$ equal to $0$ to find where the expression is undefined.

$3 x - 6 = 0$

Solve for $x$ .

Add $6$ to both sides of the equation.

$3 x = 6$

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

Divide each term in $3 x = 6$ by $3$ .

$\frac{3 x}{3} = \frac{6}{3}$

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 $3$ .

$x = 2$

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

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

Step 11

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 2$

$x > 2$

Step 12

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

Test a value on the interval $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$\frac{{(- 4)}^{2} - 1}{3 (- 4) - 6} > 0$

The left side $- 0.8\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 1 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - 1}{3 (0) - 6} > 0$

The left side $0.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 2$ and see if this value makes the original inequality true.

$x = 1.5$

Replace $x$ with $1.5$ in the original inequality.

$\frac{{(1.5)}^{2} - 1}{3 (1.5) - 6} > 0$

The left side $- 0.8\overline{3}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 2$ to see if it makes the inequality true.

Choose a value on the interval $x > 2$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

$\frac{{(4)}^{2} - 1}{3 (4) - 6} > 0$

The left side $2.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 2$ False

$x > 2$ True

Step 13

The solution consists of all of the true intervals.

$- 1 < x < 1$ or $x > 2$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 1 or x > 2$

Interval Notation:

$(- 1, 1) \cup (2, \infty)$

Step 15

Solve the Inequality for x (x^2-1)/(3x-6)>0

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