Precalculus Examples

$\frac{2}{x + 2} + \frac{x}{2 - x} < \frac{13}{4 - x^{2}}$

Step 1

Subtract $\frac{13}{4 - x^{2}}$ from both sides of the inequality.

$\frac{2}{x + 2} + \frac{x}{2 - x} - \frac{13}{4 - x^{2}} < 0$

Step 2

Simplify $\frac{2}{x + 2} + \frac{x}{2 - x} - \frac{13}{4 - x^{2}}$ .

Tap for more steps...

Step 2.1

Simplify the denominator.

Tap for more steps...

Step 2.1.1

Rewrite $4$ as $2^{2}$ .

$\frac{2}{x + 2} + \frac{x}{2 - x} - \frac{13}{2^{2} - x^{2}} < 0$

Step 2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$\frac{2}{x + 2} + \frac{x}{2 - x} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.2

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

$\frac{2}{x + 2} \cdot \frac{2 - x}{2 - x} + \frac{x}{2 - x} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.3

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

$\frac{2}{x + 2} \cdot \frac{2 - x}{2 - x} + \frac{x}{2 - x} \cdot \frac{x + 2}{x + 2} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.4

Write each expression with a common denominator of $(x + 2) (2 - x)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 2.4.1

Multiply $\frac{2}{x + 2}$ by $\frac{2 - x}{2 - x}$ .

$\frac{2 (2 - x)}{(x + 2) (2 - x)} + \frac{x}{2 - x} \cdot \frac{x + 2}{x + 2} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.4.2

Multiply $\frac{x}{2 - x}$ by $\frac{x + 2}{x + 2}$ .

$\frac{2 (2 - x)}{(x + 2) (2 - x)} + \frac{x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.4.3

Reorder the factors of $(x + 2) (2 - x)$ .

$\frac{2 (2 - x)}{(2 - x) (x + 2)} + \frac{x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{2 (2 - x) + x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6

Simplify the numerator.

Tap for more steps...

Step 2.6.1

Apply the distributive property.

$\frac{2 \cdot 2 + 2 (- x) + x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.2

Multiply $2$ by $2$ .

$\frac{4 + 2 (- x) + x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.3

Multiply $- 1$ by $2$ .

$\frac{4 - 2 x + x (x + 2)}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.4

Apply the distributive property.

$\frac{4 - 2 x + x \cdot x + x \cdot 2}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.5

Multiply $x$ by $x$ .

$\frac{4 - 2 x + x^{2} + x \cdot 2}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.6

Move $2$ to the left of $x$ .

$\frac{4 - 2 x + x^{2} + 2 \cdot x}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.7

Add $- 2 x$ and $2 x$ .

$\frac{4 + x^{2} + 0}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.6.8

Add $4 + x^{2}$ and $0$ .

$\frac{4 + x^{2}}{(2 - x) (x + 2)} - \frac{13}{(2 + x) (2 - x)} < 0$

Step 2.7

Reorder terms.

$\frac{4 + x^{2}}{(2 - x) (x + 2)} - \frac{13}{(x + 2) (2 - x)} < 0$

Step 2.8

Reorder the factors of $(x + 2) (2 - x)$ .

$\frac{4 + x^{2}}{(2 - x) (x + 2)} - \frac{13}{(2 - x) (x + 2)} < 0$

Step 2.9

Combine the numerators over the common denominator.

$\frac{4 + x^{2} - 13}{(2 - x) (x + 2)} < 0$

Step 2.10

Simplify the numerator.

Tap for more steps...

Step 2.10.1

Subtract $13$ from $4$ .

$\frac{x^{2} - 9}{(2 - x) (x + 2)} < 0$

Step 2.10.2

Rewrite $9$ as $3^{2}$ .

$\frac{x^{2} - 3^{2}}{(2 - x) (x + 2)} < 0$

Step 2.10.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

$\frac{(x + 3) (x - 3)}{(2 - x) (x + 2)} < 0$

Step 3

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

$x + 3 = 0$

$x - 3 = 0$

$2 - x = 0$

$x + 2 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 7

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 7.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 7.2

Simplify the left side.

Tap for more steps...

Step 7.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 7.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 7.3

Simplify the right side.

Tap for more steps...

Step 7.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 8

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 9

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

$x = - 3$

$x = 3$

$x = 2$

$x = - 2$

Step 10

Consolidate the solutions.

$x = - 3, 3, 2, - 2$

Step 11

Find the domain of $\frac{(x + 3) (x - 3)}{(2 - x) (x + 2)}$ .

Tap for more steps...

Step 11.1

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

$(2 - x) (x + 2) = 0$

Step 11.2

Solve for $x$ .

Tap for more steps...

Step 11.2.1

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

$2 - x = 0$

$x + 2 = 0$

Step 11.2.2

Set $2 - x$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 11.2.2.1

Set $2 - x$ equal to $0$ .

$2 - x = 0$

Step 11.2.2.2

Solve $2 - x = 0$ for $x$ .

Tap for more steps...

Step 11.2.2.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 11.2.2.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 11.2.2.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 11.2.2.2.2.2

Simplify the left side.

Tap for more steps...

Step 11.2.2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 11.2.2.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 11.2.2.2.2.3

Simplify the right side.

Tap for more steps...

Step 11.2.2.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 11.2.3

Set $x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 11.2.3.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 11.2.3.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 11.2.4

The final solution is all the values that make $(2 - x) (x + 2) = 0$ true.

$x = 2, - 2$

Step 11.3

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

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Step 12

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - 2$

$- 2 < x < 2$

$2 < x < 3$

$x > 3$

Step 13

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 13.1

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

Tap for more steps...

Step 13.1.1

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

$x = - 6$

Step 13.1.2

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

$\frac{2}{(- 6) + 2} + \frac{- 6}{2 - (- 6)} < \frac{13}{4 - {(- 6)}^{2}}$

Step 13.1.3

The left side $- 1.25$ is less than the right side $- 0.40625$ , which means that the given statement is always true.

True

Step 13.2

Test a value on the interval $- 3 < x < - 2$ to see if it makes the inequality true.

Tap for more steps...

Step 13.2.1

Choose a value on the interval $- 3 < x < - 2$ and see if this value makes the original inequality true.

$x = - 2.5$

Step 13.2.2

Replace $x$ with $- 2.5$ in the original inequality.

$\frac{2}{(- 2.5) + 2} + \frac{- 2.5}{2 - (- 2.5)} < \frac{13}{4 - {(- 2.5)}^{2}}$

Step 13.2.3

The left side $- 4.\overline{5}$ is not less than the right side $- 5.\overline{7}$ , which means that the given statement is false.

False

Step 13.3

Test a value on the interval $- 2 < x < 2$ to see if it makes the inequality true.

Tap for more steps...

Step 13.3.1

Choose a value on the interval $- 2 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 13.3.2

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

$\frac{2}{(0) + 2} + \frac{0}{2 - (0)} < \frac{13}{4 - {(0)}^{2}}$

Step 13.3.3

The left side $1$ is less than the right side $3.25$ , which means that the given statement is always true.

True

Step 13.4

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Tap for more steps...

Step 13.4.1

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Step 13.4.2

Replace $x$ with $2.5$ in the original inequality.

$\frac{2}{(2.5) + 2} + \frac{2.5}{2 - (2.5)} < \frac{13}{4 - {(2.5)}^{2}}$

Step 13.4.3

The left side $- 4.\overline{5}$ is not less than the right side $- 5.\overline{7}$ , which means that the given statement is false.

False

Step 13.5

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

Tap for more steps...

Step 13.5.1

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

$x = 6$

Step 13.5.2

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

$\frac{2}{(6) + 2} + \frac{6}{2 - (6)} < \frac{13}{4 - {(6)}^{2}}$

Step 13.5.3

The left side $- 1.25$ is less than the right side $- 0.40625$ , which means that the given statement is always true.

True

Step 13.6

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

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 3$ False

$x > 3$ True

$x < - 3$ True

$- 3 < x < - 2$ False

$- 2 < x < 2$ True

$2 < x < 3$ False

$x > 3$ True

Step 14

The solution consists of all of the true intervals.

$x < - 3$ or $- 2 < x < 2$ or $x > 3$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - 2 < x < 2 or x > 3$

Interval Notation:

$(- \infty, - 3) \cup (- 2, 2) \cup (3, \infty)$

Step 16