Precalculus Examples

$\frac{5 x - 7}{x^{2} - 1} \geq 0$

Step 1

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

$5 x - 7 = 0$

$x^{2} - 1 = 0$

Step 2

Add $7$ to both sides of the equation.

$5 x = 7$

Step 3

Divide each term in $5 x = 7$ by $5$ and simplify.

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Step 3.1

Divide each term in $5 x = 7$ by $5$ .

$\frac{5 x}{5} = \frac{7}{5}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $5$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{7}{5}$

Step 3.2.1.2

Divide $x$ by $1$ .

$x = \frac{7}{5}$

Step 4

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 5

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

$x = \pm \sqrt{1}$

Step 6

Any root of $1$ is $1$ .

$x = \pm 1$

Step 7

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

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Step 7.1

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

$x = 1$

Step 7.2

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

$x = - 1$

Step 7.3

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

$x = 1, - 1$

Step 8

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

$x = \frac{7}{5}$

$x = 1, - 1$

Step 9

Consolidate the solutions.

$x = \frac{7}{5}, 1, - 1$

Step 10

Find the domain of $\frac{5 x - 7}{x^{2} - 1}$ .

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Step 10.1

Set the denominator in $\frac{5 x - 7}{x^{2} - 1}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 1 = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 10.2.2

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

$x = \pm \sqrt{1}$

Step 10.2.3

Any root of $1$ is $1$ .

$x = \pm 1$

Step 10.2.4

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

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Step 10.2.4.1

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

$x = 1$

Step 10.2.4.2

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

$x = - 1$

Step 10.2.4.3

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

$x = 1, - 1$

Step 10.3

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

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

Step 11

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < \frac{7}{5}$

$x > \frac{7}{5}$

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 $x < - 1$ to see if it makes the inequality true.

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Step 12.1.1

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

$x = - 4$

Step 12.1.2

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

$\frac{5 (- 4) - 7}{{(- 4)}^{2} - 1} \geq 0$

Step 12.1.3

The left side $- 1.8$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.2

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

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Step 12.2.1

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

$x = 0$

Step 12.2.2

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

$\frac{5 (0) - 7}{{(0)}^{2} - 1} \geq 0$

Step 12.2.3

The left side $7$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $1 < x < \frac{7}{5}$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $1 < x < \frac{7}{5}$ and see if this value makes the original inequality true.

$x = 1.2$

Step 12.3.2

Replace $x$ with $1.2$ in the original inequality.

$\frac{5 (1.2) - 7}{{(1.2)}^{2} - 1} \geq 0$

Step 12.3.3

The left side $- 2.\overline{27}$ is less than the right side $0$ , which means that the given statement is false.

False

Step 12.4

Test a value on the interval $x > \frac{7}{5}$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $x > \frac{7}{5}$ and see if this value makes the original inequality true.

$x = 4$

Step 12.4.2

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

$\frac{5 (4) - 7}{{(4)}^{2} - 1} \geq 0$

Step 12.4.3

The left side $0.8\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 12.5

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

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < \frac{7}{5}$ False

$x > \frac{7}{5}$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < \frac{7}{5}$ False

$x > \frac{7}{5}$ True

Step 13

The solution consists of all of the true intervals.

$- 1 < x < 1$ or $x \geq \frac{7}{5}$

Step 14

Convert the inequality to interval notation.

$(- 1, 1) \cup [\frac{7}{5}, \infty)$

Step 15