Precalculus Examples

$x (3 x + 1) > 0$ , $x - 2 < 100$

Step 1

Simplify the first inequality.

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Step 1.1

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

$x = 0$

$3 x + 1 = 0$ or $x - 2 < 100$

Step 1.2

Set $x$ equal to $0$ .

$x = 0$ or $x - 2 < 100$

Step 1.3

Set $3 x + 1$ equal to $0$ and solve for $x$ .

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Step 1.3.1

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$ or $x - 2 < 100$

Step 1.3.2

Solve $3 x + 1 = 0$ for $x$ .

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Step 1.3.2.1

Subtract $1$ from both sides of the equation.

$3 x = - 1$ or $x - 2 < 100$

Step 1.3.2.2

Divide each term in $3 x = - 1$ by $3$ and simplify.

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Step 1.3.2.2.1

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$ or $x - 2 < 100$

Step 1.3.2.2.2

Simplify the left side.

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Step 1.3.2.2.2.1

Cancel the common factor of $3$ .

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Step 1.3.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$ or $x - 2 < 100$

Step 1.3.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$ or $x - 2 < 100$

Step 1.3.2.2.3

Simplify the right side.

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Step 1.3.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{3}$ or $x - 2 < 100$

Step 1.4

The final solution is all the values that make $x (3 x + 1) > 0$ true.

$x = 0, - \frac{1}{3}$ or $x - 2 < 100$

Step 1.5

Use each root to create test intervals.

$x < - \frac{1}{3}$

$- \frac{1}{3} < x < 0$

$x > 0$ or $x - 2 < 100$

Step 1.6

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 1.6.1

Test a value on the interval $x < - \frac{1}{3}$ to see if it makes the inequality true.

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Step 1.6.1.1

Choose a value on the interval $x < - \frac{1}{3}$ and see if this value makes the original inequality true.

$x = - 3$ or $x - 2 < 100$

Step 1.6.1.2

Replace $x$ with $- 3$ in the original inequality.

$(- 3) (3 (- 3) + 1) > 0$ or $x - 2 < 100$

Step 1.6.1.3

The left side $24$ is greater than the right side $0$ , which means that the given statement is always true.

True or $x - 2 < 100$

Step 1.6.2

Test a value on the interval $- \frac{1}{3} < x < 0$ to see if it makes the inequality true.

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Step 1.6.2.1

Choose a value on the interval $- \frac{1}{3} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.17$ or $x - 2 < 100$

Step 1.6.2.2

Replace $x$ with $- 0.17$ in the original inequality.

$(- 0.17) (3 (- 0.17) + 1) > 0$ or $x - 2 < 100$

Step 1.6.2.3

The left side $- 0.0833$ is not greater than the right side $0$ , which means that the given statement is false.

False or $x - 2 < 100$

Step 1.6.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 1.6.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$ or $x - 2 < 100$

Step 1.6.3.2

Replace $x$ with $2$ in the original inequality.

$(2) (3 (2) + 1) > 0$ or $x - 2 < 100$

Step 1.6.3.3

The left side $14$ is greater than the right side $0$ , which means that the given statement is always true.

True or $x - 2 < 100$

Step 1.6.4

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

$x < - \frac{1}{3}$ True

$- \frac{1}{3} < x < 0$ False

$x > 0$ True or $x - 2 < 100$

$x < - \frac{1}{3}$ True

$- \frac{1}{3} < x < 0$ False

$x > 0$ True or $x - 2 < 100$

Step 1.7

The solution consists of all of the true intervals.

$x < - \frac{1}{3}$ or $x > 0$ or $x - 2 < 100$

Step 2

Move all terms not containing $x$ to the right side of the inequality.

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Step 2.1

Add $2$ to both sides of the inequality.

$x < - \frac{1}{3}$ or $x > 0$ or $x < 100 + 2$

Step 2.2

Add $100$ and $2$ .

$x < - \frac{1}{3}$ or $x > 0$ or $x < 102$

Step 3

The union consists of all of the elements that are contained in each interval.

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Step 5