Precalculus Examples

$\frac{3}{x - 1} > \frac{4}{x + 1}$

Step 1

Subtract $\frac{4}{x + 1}$ from both sides of the inequality.

$\frac{3}{x - 1} - \frac{4}{x + 1} > 0$

Step 2

Simplify $\frac{3}{x - 1} - \frac{4}{x + 1}$ .

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Step 2.1

To write $\frac{3}{x - 1}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{3}{x - 1} \cdot \frac{x + 1}{x + 1} - \frac{4}{x + 1} > 0$

Step 2.2

To write $- \frac{4}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{3}{x - 1} \cdot \frac{x + 1}{x + 1} - \frac{4}{x + 1} \cdot \frac{x - 1}{x - 1} > 0$

Step 2.3

Write each expression with a common denominator of $(x - 1) (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

Multiply $\frac{3}{x - 1}$ by $\frac{x + 1}{x + 1}$ .

$\frac{3 (x + 1)}{(x - 1) (x + 1)} - \frac{4}{x + 1} \cdot \frac{x - 1}{x - 1} > 0$

Step 2.3.2

Multiply $\frac{4}{x + 1}$ by $\frac{x - 1}{x - 1}$ .

$\frac{3 (x + 1)}{(x - 1) (x + 1)} - \frac{4 (x - 1)}{(x + 1) (x - 1)} > 0$

Step 2.3.3

Reorder the factors of $(x - 1) (x + 1)$ .

$\frac{3 (x + 1)}{(x + 1) (x - 1)} - \frac{4 (x - 1)}{(x + 1) (x - 1)} > 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{3 (x + 1) - 4 (x - 1)}{(x + 1) (x - 1)} > 0$

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

$\frac{3 x + 3 \cdot 1 - 4 (x - 1)}{(x + 1) (x - 1)} > 0$

Step 2.5.2

Multiply $3$ by $1$ .

$\frac{3 x + 3 - 4 (x - 1)}{(x + 1) (x - 1)} > 0$

Step 2.5.3

Apply the distributive property.

$\frac{3 x + 3 - 4 x - 4 \cdot - 1}{(x + 1) (x - 1)} > 0$

Step 2.5.4

Multiply $- 4$ by $- 1$ .

$\frac{3 x + 3 - 4 x + 4}{(x + 1) (x - 1)} > 0$

Step 2.5.5

Subtract $4 x$ from $3 x$ .

$\frac{- x + 3 + 4}{(x + 1) (x - 1)} > 0$

Step 2.5.6

Add $3$ and $4$ .

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

Step 2.6

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

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

Step 2.7

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

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

Step 2.8

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

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

Step 2.9

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

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

Step 2.10

Move the negative in front of the fraction.

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

Step 3

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

$x - 7 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 4

Add $7$ to both sides of the equation.

$x = 7$

Step 5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 6

Add $1$ to both sides of the equation.

$x = 1$

Step 7

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

$x = 7$

$x = - 1$

$x = 1$

Step 8

Consolidate the solutions.

$x = 7, - 1, 1$

Step 9

Find the domain of $- \frac{x - 7}{(x + 1) (x - 1)}$ .

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Step 9.1

Set the denominator in $\frac{x - 7}{(x + 1) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x + 1) (x - 1) = 0$

Step 9.2

Solve for $x$ .

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Step 9.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x - 1 = 0$

Step 9.2.2

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 9.2.2.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 9.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 9.2.3

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 9.2.3.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Step 9.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 9.2.4

The final solution is all the values that make $(x + 1) (x - 1) = 0$ true.

$x = - 1, 1$

Step 9.3

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

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

Step 10

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 7$

$x > 7$

Step 11

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 11.1

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

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Step 11.1.1

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

$x = - 4$

Step 11.1.2

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

$\frac{3}{(- 4) - 1} > \frac{4}{(- 4) + 1}$

Step 11.1.3

The left side $- 0.6$ is greater than the right side $- 1.\overline{3}$ , which means that the given statement is always true.

True

Step 11.2

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

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Step 11.2.1

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

$x = 0$

Step 11.2.2

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

$\frac{3}{(0) - 1} > \frac{4}{(0) + 1}$

Step 11.2.3

The left side $- 3$ is not greater than the right side $4$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $1 < x < 7$ to see if it makes the inequality true.

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Step 11.3.1

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

$x = 4$

Step 11.3.2

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

$\frac{3}{(4) - 1} > \frac{4}{(4) + 1}$

Step 11.3.3

The left side $1$ is greater than the right side $0.8$ , which means that the given statement is always true.

True

Step 11.4

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

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Step 11.4.1

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

$x = 10$

Step 11.4.2

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

$\frac{3}{(10) - 1} > \frac{4}{(10) + 1}$

Step 11.4.3

The left side $0.\overline{3}$ is not greater than the right side $0.\overline{36}$ , which means that the given statement is false.

False

Step 11.5

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

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 7$ True

$x > 7$ False

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 7$ True

$x > 7$ False

Step 12

The solution consists of all of the true intervals.

$x < - 1$ or $1 < x < 7$

Step 13

Convert the inequality to interval notation.

$(- \infty, - 1) \cup (1, 7)$

Step 14