Finite Math Examples

$\frac{x + 1}{x - 1} + 2 > 0$

Step 1

Simplify $\frac{x + 1}{x - 1} + 2$ .

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Step 1.1

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

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

Step 1.2

Combine the numerators over the common denominator.

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

Step 1.3

Simplify the numerator.

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Step 1.3.1

Apply the distributive property.

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

Step 1.3.2

Multiply $2$ by $- 1$ .

$\frac{x + 1 + 2 x - 2}{x - 1} > 0$

Step 1.3.3

Add $x$ and $2 x$ .

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

Step 1.3.4

Subtract $2$ from $1$ .

$\frac{3 x - 1}{x - 1} > 0$

Step 2

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

$3 x - 1 = 0$

$x - 1 = 0$

Step 3

Add $1$ to both sides of the equation.

$3 x = 1$

Step 4

Divide each term in $3 x = 1$ by $3$ and simplify.

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Step 4.1

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $3$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 5

Add $1$ to both sides of the equation.

$x = 1$

Step 6

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

$x = \frac{1}{3}$

$x = 1$

Step 7

Consolidate the solutions.

$x = \frac{1}{3}, 1$

Step 8

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

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Step 8.1

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

$x - 1 = 0$

Step 8.2

Add $1$ to both sides of the equation.

$x = 1$

Step 8.3

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

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

Step 9

Use each root to create test intervals.

$x < \frac{1}{3}$

$\frac{1}{3} < x < 1$

$x > 1$

Step 10

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 10.1

Test a value on the interval $x < \frac{1}{3}$ to see if it makes the inequality true.

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Step 10.1.1

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

$x = 0$

Step 10.1.2

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

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

Step 10.1.3

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

True

Step 10.2

Test a value on the interval $\frac{1}{3} < x < 1$ to see if it makes the inequality true.

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Step 10.2.1

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

$x = 0.67$

Step 10.2.2

Replace $x$ with $0.67$ in the original inequality.

$\frac{(0.67) + 1}{(0.67) - 1} + 2 > 0$

Step 10.2.3

The left side $- 3.\overline{06}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 10.3

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

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Step 10.3.1

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

$x = 4$

Step 10.3.2

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

$\frac{(4) + 1}{(4) - 1} + 2 > 0$

Step 10.3.3

The left side $3.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.4

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

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 1$ False

$x > 1$ True

$x < \frac{1}{3}$ True

$\frac{1}{3} < x < 1$ False

$x > 1$ True

Step 11

The solution consists of all of the true intervals.

$x < \frac{1}{3}$ or $x > 1$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$x < \frac{1}{3} or x > 1$

Interval Notation:

$(- \infty, \frac{1}{3}) \cup (1, \infty)$

Step 13