Algebra Examples

\frac{4 x + 5}{x} > 3

$\frac{4 x + 5}{x} > 3$

Step 1

Subtract

3

$3$ from both sides of the inequality.

\frac{4 x + 5}{x} - 3 > 0

$\frac{4 x + 5}{x} - 3 > 0$

Step 2

Simplify

\frac{4 x + 5}{x} - 3

$\frac{4 x + 5}{x} - 3$ .

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Step 2.1

To write

- 3

$- 3$ as a fraction with a common denominator, multiply by

\frac{x}{x}

$\frac{x}{x}$ .

\frac{4 x + 5}{x} - 3 \cdot \frac{x}{x} > 0

$\frac{4 x + 5}{x} - 3 \cdot \frac{x}{x} > 0$

Step 2.2

Combine

- 3

$- 3$ and

\frac{x}{x}

$\frac{x}{x}$ .

\frac{4 x + 5}{x} + \frac{- 3 x}{x} > 0

$\frac{4 x + 5}{x} + \frac{- 3 x}{x} > 0$

Step 2.3

Combine the numerators over the common denominator.

\frac{4 x + 5 - 3 x}{x} > 0

$\frac{4 x + 5 - 3 x}{x} > 0$

Step 2.4

Subtract

3 x

$3 x$ from

4 x

$4 x$ .

\frac{x + 5}{x} > 0

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

\frac{x + 5}{x} > 0

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

Step 3

Find all the values where the expression switches from negative to positive by setting each factor equal to

0

$0$ and solving.

x = 0

$x = 0$

x + 5 = 0

$x + 5 = 0$

Step 4

Subtract

5

$5$ from both sides of the equation.

x = - 5

$x = - 5$

Step 5

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

x = 0

$x = 0$

x = - 5

$x = - 5$

Step 6

Consolidate the solutions.

x = 0, - 5

$x = 0, - 5$

Step 7

Find the domain of

\frac{x + 5}{x}

$\frac{x + 5}{x}$ .

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Step 7.1

Set the denominator in

\frac{x + 5}{x}

$\frac{x + 5}{x}$ equal to

0

$0$ to find where the expression is undefined.

x = 0

$x = 0$

Step 7.2

The domain is all values of

x

$x$ that make the expression defined.

(- \infty, 0) \cup (0, \infty)

$(- \infty, 0) \cup (0, \infty)$

(- \infty, 0) \cup (0, \infty)

$(- \infty, 0) \cup (0, \infty)$

Step 8

Use each root to create test intervals.

x < - 5

$x < - 5$

- 5 < x < 0

$- 5 < x < 0$

x > 0

$x > 0$

Step 9

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 9.1

Test a value on the interval

x < - 5

$x < - 5$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval

x < - 5

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

x = - 8

$x = - 8$

Step 9.1.2

Replace

x

$x$ with

- 8

$- 8$ in the original inequality.

\frac{4 (- 8) + 5}{- 8} > 3

$\frac{4 (- 8) + 5}{- 8} > 3$

Step 9.1.3

The left side

3.375

$3.375$ is greater than the right side

3

$3$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval

- 5 < x < 0

$- 5 < x < 0$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval

- 5 < x < 0

$- 5 < x < 0$ and see if this value makes the original inequality true.

x = - 2

$x = - 2$

Step 9.2.2

Replace

x

$x$ with

- 2

$- 2$ in the original inequality.

\frac{4 (- 2) + 5}{- 2} > 3

$\frac{4 (- 2) + 5}{- 2} > 3$

Step 9.2.3

The left side

1.5

$1.5$ is not greater than the right side

3

$3$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval

x > 0

$x > 0$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval

x > 0

$x > 0$ and see if this value makes the original inequality true.

x = 2

$x = 2$

Step 9.3.2

Replace

x

$x$ with

2

$2$ in the original inequality.

\frac{4 (2) + 5}{2} > 3

$\frac{4 (2) + 5}{2} > 3$

Step 9.3.3

The left side

6.5

$6.5$ is greater than the right side

3

$3$ , which means that the given statement is always true.

True

Step 9.4

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

x < - 5

$x < - 5$ True

- 5 < x < 0

$- 5 < x < 0$ False

x > 0

$x > 0$ True

x < - 5

$x < - 5$ True

- 5 < x < 0

$- 5 < x < 0$ False

x > 0

$x > 0$ True

Step 10

The solution consists of all of the true intervals.

x < - 5

$x < - 5$ or

x > 0

$x > 0$

Step 11

The result can be shown in multiple forms.

Inequality Form:

x < - 5 or x > 0

$x < - 5 or x > 0$

Interval Notation:

(- \infty, - 5) \cup (0, \infty)

$(- \infty, - 5) \cup (0, \infty)$

Step 12

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