Precalculus Examples

$\frac{x - 4}{x + 11} \geq \frac{1}{x - 1}$

Step 1

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

$\frac{x - 4}{x + 11} - \frac{1}{x - 1} \geq 0$

Step 2

Simplify $\frac{x - 4}{x + 11} - \frac{1}{x - 1}$ .

Tap for more steps...

Step 2.1

To write $\frac{x - 4}{x + 11}$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

$\frac{x - 4}{x + 11} \cdot \frac{x - 1}{x - 1} - \frac{1}{x - 1} \geq 0$

Step 2.2

To write $- \frac{1}{x - 1}$ as a fraction with a common denominator, multiply by $\frac{x + 11}{x + 11}$ .

$\frac{x - 4}{x + 11} \cdot \frac{x - 1}{x - 1} - \frac{1}{x - 1} \cdot \frac{x + 11}{x + 11} \geq 0$

Step 2.3

Write each expression with a common denominator of $(x + 11) (x - 1)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 2.3.1

Multiply $\frac{x - 4}{x + 11}$ by $\frac{x - 1}{x - 1}$ .

$\frac{(x - 4) (x - 1)}{(x + 11) (x - 1)} - \frac{1}{x - 1} \cdot \frac{x + 11}{x + 11} \geq 0$

Step 2.3.2

Multiply $\frac{1}{x - 1}$ by $\frac{x + 11}{x + 11}$ .

$\frac{(x - 4) (x - 1)}{(x + 11) (x - 1)} - \frac{x + 11}{(x - 1) (x + 11)} \geq 0$

Step 2.3.3

Reorder the factors of $(x - 1) (x + 11)$ .

$\frac{(x - 4) (x - 1)}{(x + 11) (x - 1)} - \frac{x + 11}{(x + 11) (x - 1)} \geq 0$

Step 2.4

Combine the numerators over the common denominator.

$\frac{(x - 4) (x - 1) - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5

Simplify the numerator.

Tap for more steps...

Step 2.5.1

Expand $(x - 4) (x - 1)$ using the FOIL Method.

Tap for more steps...

Step 2.5.1.1

Apply the distributive property.

$\frac{x (x - 1) - 4 (x - 1) - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.1.2

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 1 - 4 (x - 1) - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.1.3

Apply the distributive property.

$\frac{x \cdot x + x \cdot - 1 - 4 x - 4 \cdot - 1 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.2

Simplify and combine like terms.

Tap for more steps...

Step 2.5.2.1

Simplify each term.

Tap for more steps...

Step 2.5.2.1.1

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot - 1 - 4 x - 4 \cdot - 1 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.2.1.2

Move $- 1$ to the left of $x$ .

$\frac{x^{2} - 1 \cdot x - 4 x - 4 \cdot - 1 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.2.1.3

Rewrite $- 1 x$ as $- x$ .

$\frac{x^{2} - x - 4 x - 4 \cdot - 1 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.2.1.4

Multiply $- 4$ by $- 1$ .

$\frac{x^{2} - x - 4 x + 4 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.2.2

Subtract $4 x$ from $- x$ .

$\frac{x^{2} - 5 x + 4 - (x + 11)}{(x + 11) (x - 1)} \geq 0$

Step 2.5.3

Apply the distributive property.

$\frac{x^{2} - 5 x + 4 - x - 1 \cdot 11}{(x + 11) (x - 1)} \geq 0$

Step 2.5.4

Multiply $- 1$ by $11$ .

$\frac{x^{2} - 5 x + 4 - x - 11}{(x + 11) (x - 1)} \geq 0$

Step 2.5.5

Subtract $x$ from $- 5 x$ .

$\frac{x^{2} - 6 x + 4 - 11}{(x + 11) (x - 1)} \geq 0$

Step 2.5.6

Subtract $11$ from $4$ .

$\frac{x^{2} - 6 x - 7}{(x + 11) (x - 1)} \geq 0$

Step 2.5.7

Factor $x^{2} - 6 x - 7$ using the AC method.

Tap for more steps...

Step 2.5.7.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 7$ and whose sum is $- 6$ .

$- 7, 1$

Step 2.5.7.2

Write the factored form using these integers.

$\frac{(x - 7) (x + 1)}{(x + 11) (x - 1)} \geq 0$

Step 3

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

$x - 7 = 0$

$x + 1 = 0$

$x + 11 = 0$

$x - 1 = 0$

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

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 7

Add $1$ to both sides of the equation.

$x = 1$

Step 8

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

$x = 7$

$x = - 1$

$x = - 11$

$x = 1$

Step 9

Consolidate the solutions.

$x = 7, - 1, - 11, 1$

Step 10

Find the domain of $\frac{(x - 7) (x + 1)}{(x + 11) (x - 1)}$ .

Tap for more steps...

Step 10.1

Set the denominator in $\frac{(x - 7) (x + 1)}{(x + 11) (x - 1)}$ equal to $0$ to find where the expression is undefined.

$(x + 11) (x - 1) = 0$

Step 10.2

Solve for $x$ .

Tap for more steps...

Step 10.2.1

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

$x + 11 = 0$

$x - 1 = 0$

Step 10.2.2

Set $x + 11$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 10.2.2.1

Set $x + 11$ equal to $0$ .

$x + 11 = 0$

Step 10.2.2.2

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 10.2.3

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

Tap for more steps...

Step 10.2.3.1

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

$x - 1 = 0$

Step 10.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 10.2.4

The final solution is all the values that make $(x + 11) (x - 1) = 0$ true.

$x = - 11, 1$

Step 10.3

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

$(- \infty, - 11) \cup (- 11, 1) \cup (1, \infty)$

Step 11

Use each root to create test intervals.

$x < - 11$

$- 11 < x < - 1$

$- 1 < x < 1$

$1 < x < 7$

$x > 7$

Step 12

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

Tap for more steps...

Step 12.1

Test a value on the interval $x < - 11$ to see if it makes the inequality true.

Tap for more steps...

Step 12.1.1

Choose a value on the interval $x < - 11$ and see if this value makes the original inequality true.

$x = - 14$

Step 12.1.2

Replace $x$ with $- 14$ in the original inequality.

$\frac{(- 14) - 4}{(- 14) + 11} \geq \frac{1}{(- 14) - 1}$

Step 12.1.3

The left side $6$ is greater than the right side $- 0.0\overline{6}$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $- 11 < x < - 1$ to see if it makes the inequality true.

Tap for more steps...

Step 12.2.1

Choose a value on the interval $- 11 < x < - 1$ and see if this value makes the original inequality true.

$x = - 6$

Step 12.2.2

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

$\frac{(- 6) - 4}{(- 6) + 11} \geq \frac{1}{(- 6) - 1}$

Step 12.2.3

The left side $- 2$ is less than the right side $- 0.\overline{142857}$ , 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.

Tap for more steps...

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.

$\frac{(0) - 4}{(0) + 11} \geq \frac{1}{(0) - 1}$

Step 12.3.3

The left side $- 0.\overline{36}$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 12.4

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

Tap for more steps...

Step 12.4.1

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

$x = 4$

Step 12.4.2

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

$\frac{(4) - 4}{(4) + 11} \geq \frac{1}{(4) - 1}$

Step 12.4.3

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

False

Step 12.5

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

Tap for more steps...

Step 12.5.1

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

$x = 10$

Step 12.5.2

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

$\frac{(10) - 4}{(10) + 11} \geq \frac{1}{(10) - 1}$

Step 12.5.3

The left side $0.\overline{285714}$ is greater than the right side $0.\overline{1}$ , which means that the given statement is always true.

True

Step 12.6

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

$x < - 11$ True

$- 11 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 7$ False

$x > 7$ True

$x < - 11$ True

$- 11 < x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 7$ False

$x > 7$ True

Step 13

The solution consists of all of the true intervals.

$x < - 11$ or $- 1 \leq x < 1$ or $x \geq 7$

Step 14

Convert the inequality to interval notation.

$(- \infty, - 11) \cup [- 1, 1) \cup [7, \infty)$

Step 15