Precalculus Examples

\frac{2 x + 6}{x} < 1

$\frac{2 x + 6}{x} < 1$

Step 1

Subtract

1

$1$ from both sides of the inequality.

\frac{2 x + 6}{x} - 1 < 0

$\frac{2 x + 6}{x} - 1 < 0$

Step 2

Simplify

\frac{2 x + 6}{x} - 1

$\frac{2 x + 6}{x} - 1$ .

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Step 2.1

Factor

2

$2$ out of

2 x + 6

$2 x + 6$ .

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Step 2.1.1

Factor

2

$2$ out of

2 x

$2 x$ .

\frac{2 (x) + 6}{x} - 1 < 0

$\frac{2 (x) + 6}{x} - 1 < 0$

Step 2.1.2

Factor

2

$2$ out of

6

$6$ .

\frac{2 x + 2 \cdot 3}{x} - 1 < 0

$\frac{2 x + 2 \cdot 3}{x} - 1 < 0$

Step 2.1.3

Factor

2

$2$ out of

2 x + 2 \cdot 3

$2 x + 2 \cdot 3$ .

\frac{2 (x + 3)}{x} - 1 < 0

$\frac{2 (x + 3)}{x} - 1 < 0$

\frac{2 (x + 3)}{x} - 1 < 0

$\frac{2 (x + 3)}{x} - 1 < 0$

Step 2.2

To write

- 1

$- 1$ as a fraction with a common denominator, multiply by

\frac{x}{x}

$\frac{x}{x}$ .

\frac{2 (x + 3)}{x} - 1 \cdot \frac{x}{x} < 0

$\frac{2 (x + 3)}{x} - 1 \cdot \frac{x}{x} < 0$

Step 2.3

Combine

- 1

$- 1$ and

\frac{x}{x}

$\frac{x}{x}$ .

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

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

Step 2.4

Combine the numerators over the common denominator.

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

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

\frac{2 x + 2 \cdot 3 - x}{x} < 0

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

Step 2.5.2

Multiply

2

$2$ by

3

$3$ .

\frac{2 x + 6 - x}{x} < 0

$\frac{2 x + 6 - x}{x} < 0$

Step 2.5.3

Subtract

x

$x$ from

2 x

$2 x$ .

\frac{x + 6}{x} < 0

$\frac{x + 6}{x} < 0$

\frac{x + 6}{x} < 0

$\frac{x + 6}{x} < 0$

\frac{x + 6}{x} < 0

$\frac{x + 6}{x} < 0$

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 + 6 = 0

$x + 6 = 0$

Step 4

Subtract

6

$6$ from both sides of the equation.

x = - 6

$x = - 6$

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 = - 6

$x = - 6$

Step 6

Consolidate the solutions.

x = 0, - 6

$x = 0, - 6$

Step 7

Find the domain of

\frac{x + 6}{x}

$\frac{x + 6}{x}$ .

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Step 7.1

Set the denominator in

\frac{x + 6}{x}

$\frac{x + 6}{x}$ 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 < - 6

$x < - 6$

- 6 < x < 0

$- 6 < x < 0$

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 < - 6

$x < - 6$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval

x < - 6

$x < - 6$ 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{2 (- 8) + 6}{- 8} < 1

$\frac{2 (- 8) + 6}{- 8} < 1$

Step 9.1.3

The left side

1.25

$1.25$ is not less than the right side

1

$1$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval

- 6 < x < 0

$- 6 < x < 0$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval

- 6 < x < 0

$- 6 < x < 0$ and see if this value makes the original inequality true.

x = - 3

$x = - 3$

Step 9.2.2

Replace

x

$x$ with

- 3

$- 3$ in the original inequality.

\frac{2 (- 3) + 6}{- 3} < 1

$\frac{2 (- 3) + 6}{- 3} < 1$

Step 9.2.3

The left side

0

$0$ is less than the right side

1

$1$ , which means that the given statement is always true.

True

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{2 (2) + 6}{2} < 1

$\frac{2 (2) + 6}{2} < 1$

Step 9.3.3

The left side

5

$5$ is not less than the right side

1

$1$ , which means that the given statement is false.

False

Step 9.4

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

x < - 6

$x < - 6$ False

- 6 < x < 0

$- 6 < x < 0$ True

x > 0

$x > 0$ False

x < - 6

$x < - 6$ False

- 6 < x < 0

$- 6 < x < 0$ True

x > 0

$x > 0$ False

Step 10

The solution consists of all of the true intervals.

- 6 < x < 0

$- 6 < x < 0$

Step 11

The result can be shown in multiple forms.

Inequality Form:

- 6 < x < 0

$- 6 < x < 0$

Interval Notation:

(- 6, 0)

$(- 6, 0)$

Step 12

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