Precalculus Examples

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

Step 1

Subtract $\frac{1}{x - 3}$ from both sides of the inequality.

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

Step 2

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

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Step 2.1

To write $\frac{1}{x + 4}$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

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

Step 2.2

To write $- \frac{1}{x - 3}$ as a fraction with a common denominator, multiply by $\frac{x + 4}{x + 4}$ .

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

Step 2.3

Write each expression with a common denominator of $(x + 4) (x - 3)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Reorder the factors of $(x - 3) (x + 4)$ .

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

Step 2.4

Combine the numerators over the common denominator.

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Multiply $- 1$ by $4$ .

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

Step 2.5.3

Subtract $x$ from $x$ .

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

Step 2.5.4

Subtract $3$ from $0$ .

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

Step 2.5.5

Subtract $4$ from $- 3$ .

$\frac{- 7}{(x + 4) (x - 3)} > 0$

Step 2.6

Move the negative in front of the fraction.

$- \frac{7}{(x + 4) (x - 3)} > 0$

Step 3

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

$x + 4 = 0$

$x - 3 = 0$

Step 4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

$x = - 4$

$x = 3$

Step 7

Consolidate the solutions.

$x = - 4, 3$

Step 8

Find the domain of $- \frac{7}{(x + 4) (x - 3)}$ .

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Step 8.1

Set the denominator in $\frac{7}{(x + 4) (x - 3)}$ equal to $0$ to find where the expression is undefined.

$(x + 4) (x - 3) = 0$

Step 8.2

Solve for $x$ .

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Step 8.2.1

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

$x + 4 = 0$

$x - 3 = 0$

Step 8.2.2

Set $x + 4$ equal to $0$ and solve for $x$ .

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Step 8.2.2.1

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Step 8.2.2.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 8.2.3

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 8.2.3.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 8.2.3.2

Add $3$ to both sides of the equation.

$x = 3$

Step 8.2.4

The final solution is all the values that make $(x + 4) (x - 3) = 0$ true.

$x = - 4, 3$

Step 8.3

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

$(- \infty, - 4) \cup (- 4, 3) \cup (3, \infty)$

Step 9

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 3$

$x > 3$

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 $x < - 4$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 10.1.2

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

$\frac{1}{(- 6) + 4} > \frac{1}{(- 6) - 3}$

Step 10.1.3

The left side $- 0.5$ is not greater than the right side $- 0.\overline{1}$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 4 < x < 3$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 4 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 10.2.2

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

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

Step 10.2.3

The left side $0.25$ is greater than the right side $- 0.\overline{3}$ , which means that the given statement is always true.

True

Step 10.3

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

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Step 10.3.1

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

$x = 6$

Step 10.3.2

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

$\frac{1}{(6) + 4} > \frac{1}{(6) - 3}$

Step 10.3.3

The left side $0.1$ is not greater than the right side $0.\overline{3}$ , which means that the given statement is false.

False

Step 10.4

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

$x < - 4$ False

$- 4 < x < 3$ True

$x > 3$ False

$x < - 4$ False

$- 4 < x < 3$ True

$x > 3$ False

Step 11

The solution consists of all of the true intervals.

$- 4 < x < 3$

Step 12

Convert the inequality to interval notation.

$(- 4, 3)$

Step 13