Finite Math Examples

$\frac{x - 3}{x + 5} > 0$

Step 1

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

$x - 3 = 0$

$x + 5 = 0$

Step 2

Add $3$ to both sides of the equation.

$x = 3$

Step 3

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 4

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

$x = 3$

$x = - 5$

Step 5

Consolidate the solutions.

$x = 3, - 5$

Step 6

Find the domain of $\frac{x - 3}{x + 5}$ .

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Step 6.1

Set the denominator in $\frac{x - 3}{x + 5}$ equal to $0$ to find where the expression is undefined.

$x + 5 = 0$

Step 6.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 6.3

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

$(- \infty, - 5) \cup (- 5, \infty)$

Step 7

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$x > 3$

Step 8

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 8.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 8.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 8.1.2

Replace $x$ with $- 8$ in the original inequality.

$\frac{(- 8) - 3}{(- 8) + 5} > 0$

Step 8.1.3

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

True

Step 8.2

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Step 8.2.2

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

$\frac{(0) - 3}{(0) + 5} > 0$

Step 8.2.3

The left side $- 0.6$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 8.3

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

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Step 8.3.1

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

$x = 6$

Step 8.3.2

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

$\frac{(6) - 3}{(6) + 5} > 0$

Step 8.3.3

The left side $0.\overline{27}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 8.4

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

$x < - 5$ True

$- 5 < x < 3$ False

$x > 3$ True

$x < - 5$ True

$- 5 < x < 3$ False

$x > 3$ True

Step 9

The solution consists of all of the true intervals.

$x < - 5$ or $x > 3$

Step 10

Convert the inequality to interval notation.

$(- \infty, - 5) \cup (3, \infty)$

Step 11