Algebra Examples

$\frac{x^{2} - 1}{3 x + 9} \leq 0$

Step 1

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

$x^{2} - 1 = 0$

$3 x + 9 = 0$

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.

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Step 5.1

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

$x = 1$

Step 5.2

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

$x = - 1$

Step 5.3

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

$x = 1, - 1$

Step 6

Subtract $9$ from both sides of the equation.

$3 x = - 9$

Step 7

Divide each term in $3 x = - 9$ by $3$ and simplify.

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Step 7.1

Divide each term in $3 x = - 9$ by $3$ .

$\frac{3 x}{3} = \frac{- 9}{3}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $3$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 9}{3}$

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 9}{3}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $- 9$ by $3$ .

$x = - 3$

Step 8

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

$x = 1, - 1$

$x = - 3$

Step 9

Consolidate the solutions.

$x = 1, - 1, - 3$

Step 10

Find the domain of $\frac{x^{2} - 1}{3 x + 9}$ .

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Step 10.1

Set the denominator in $\frac{x^{2} - 1}{3 x + 9}$ equal to $0$ to find where the expression is undefined.

$3 x + 9 = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

Subtract $9$ from both sides of the equation.

$3 x = - 9$

Step 10.2.2

Divide each term in $3 x = - 9$ by $3$ and simplify.

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Step 10.2.2.1

Divide each term in $3 x = - 9$ by $3$ .

$\frac{3 x}{3} = \frac{- 9}{3}$

Step 10.2.2.2

Simplify the left side.

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Step 10.2.2.2.1

Cancel the common factor of $3$ .

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Step 10.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 9}{3}$

Step 10.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 9}{3}$

Step 10.2.2.3

Simplify the right side.

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Step 10.2.2.3.1

Divide $- 9$ by $3$ .

$x = - 3$

Step 10.3

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

$(- \infty, - 3) \cup (- 3, \infty)$

Step 11

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 1$

$- 1 < x < 1$

$x > 1$

Step 12

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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Step 12.1

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

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Step 12.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 12.1.2

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

$\frac{{(- 6)}^{2} - 1}{3 (- 6) + 9} \leq 0$

Step 12.1.3

The left side $- 3.\overline{8}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $- 3 < x < - 1$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $- 3 < x < - 1$ and see if this value makes the original inequality true.

$x = - 2$

Step 12.2.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{{(- 2)}^{2} - 1}{3 (- 2) + 9} \leq 0$

Step 12.2.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 12.3

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

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Step 12.3.1

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

$x = 0$

Step 12.3.2

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

$\frac{{(0)}^{2} - 1}{3 (0) + 9} \leq 0$

Step 12.3.3

The left side $- 0.\overline{1}$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 12.4

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

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Step 12.4.1

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

$x = 4$

Step 12.4.2

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

$\frac{{(4)}^{2} - 1}{3 (4) + 9} \leq 0$

Step 12.4.3

The left side $0.\overline{714285}$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 12.5

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

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

$x < - 3$ True

$- 3 < x < - 1$ False

$- 1 < x < 1$ True

$x > 1$ False

Step 13

The solution consists of all of the true intervals.

$x < - 3$ or $- 1 \leq x \leq 1$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 1 \leq x \leq 1$

Interval Notation:

$(- \infty, - 3) \cup [- 1, 1]$

Step 15