Calculus Examples

$\frac{2 x + 4}{3 x - 2} < 1$

Step 1

Subtract $1$ from both sides of the inequality.

$\frac{2 x + 4}{3 x - 2} - 1 < 0$

Step 2

Simplify $\frac{2 x + 4}{3 x - 2} - 1$ .

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Step 2.1

Factor $2$ out of $2 x + 4$ .

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Step 2.1.1

Factor $2$ out of $2 x$ .

$\frac{2 (x) + 4}{3 x - 2} - 1 < 0$

Step 2.1.2

Factor $2$ out of $4$ .

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

Step 2.1.3

Factor $2$ out of $2 x + 2 \cdot 2$ .

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

Step 2.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3 x - 2}{3 x - 2}$ .

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

Step 2.3

Combine $- 1$ and $\frac{3 x - 2}{3 x - 2}$ .

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

Step 2.4

Combine the numerators over the common denominator.

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

Step 2.5

Simplify the numerator.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Multiply $2$ by $2$ .

$\frac{2 x + 4 - (3 x - 2)}{3 x - 2} < 0$

Step 2.5.3

Apply the distributive property.

$\frac{2 x + 4 - (3 x) - - 2}{3 x - 2} < 0$

Step 2.5.4

Multiply $3$ by $- 1$ .

$\frac{2 x + 4 - 3 x - - 2}{3 x - 2} < 0$

Step 2.5.5

Multiply $- 1$ by $- 2$ .

$\frac{2 x + 4 - 3 x + 2}{3 x - 2} < 0$

Step 2.5.6

Subtract $3 x$ from $2 x$ .

$\frac{- x + 4 + 2}{3 x - 2} < 0$

Step 2.5.7

Add $4$ and $2$ .

$\frac{- x + 6}{3 x - 2} < 0$

Step 2.6

Factor $- 1$ out of $- x$ .

$\frac{- (x) + 6}{3 x - 2} < 0$

Step 2.7

Rewrite $6$ as $- 1 (- 6)$ .

$\frac{- (x) - 1 (- 6)}{3 x - 2} < 0$

Step 2.8

Factor $- 1$ out of $- (x) - 1 (- 6)$ .

$\frac{- (x - 6)}{3 x - 2} < 0$

Step 2.9

Rewrite $- (x - 6)$ as $- 1 (x - 6)$ .

$\frac{- 1 (x - 6)}{3 x - 2} < 0$

Step 2.10

Move the negative in front of the fraction.

$- \frac{x - 6}{3 x - 2} < 0$

Step 3

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

$x - 6 = 0$

$3 x - 2 = 0$

Step 4

Add $6$ to both sides of the equation.

$x = 6$

Step 5

Add $2$ to both sides of the equation.

$3 x = 2$

Step 6

Divide each term in $3 x = 2$ by $3$ and simplify.

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Step 6.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $3$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 7

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

$x = 6$

$x = \frac{2}{3}$

Step 8

Consolidate the solutions.

$x = 6, \frac{2}{3}$

Step 9

Find the domain of $- \frac{x - 6}{3 x - 2}$ .

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Step 9.1

Set the denominator in $\frac{x - 6}{3 x - 2}$ equal to $0$ to find where the expression is undefined.

$3 x - 2 = 0$

Step 9.2

Solve for $x$ .

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Step 9.2.1

Add $2$ to both sides of the equation.

$3 x = 2$

Step 9.2.2

Divide each term in $3 x = 2$ by $3$ and simplify.

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Step 9.2.2.1

Divide each term in $3 x = 2$ by $3$ .

$\frac{3 x}{3} = \frac{2}{3}$

Step 9.2.2.2

Simplify the left side.

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Step 9.2.2.2.1

Cancel the common factor of $3$ .

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Step 9.2.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{2}{3}$

Step 9.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{3}$

Step 9.3

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

$(- \infty, \frac{2}{3}) \cup (\frac{2}{3}, \infty)$

Step 10

Use each root to create test intervals.

$x < \frac{2}{3}$

$\frac{2}{3} < x < 6$

$x > 6$

Step 11

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 11.1

Test a value on the interval $x < \frac{2}{3}$ to see if it makes the inequality true.

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Step 11.1.1

Choose a value on the interval $x < \frac{2}{3}$ and see if this value makes the original inequality true.

$x = 0$

Step 11.1.2

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

$\frac{2 (0) + 4}{3 (0) - 2} < 1$

Step 11.1.3

The left side $- 2$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 11.2

Test a value on the interval $\frac{2}{3} < x < 6$ to see if it makes the inequality true.

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Step 11.2.1

Choose a value on the interval $\frac{2}{3} < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 11.2.2

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

$\frac{2 (3) + 4}{3 (3) - 2} < 1$

Step 11.2.3

The left side $1.\overline{428571}$ is not less than the right side $1$ , which means that the given statement is false.

False

Step 11.3

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 11.3.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 11.3.2

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

$\frac{2 (8) + 4}{3 (8) - 2} < 1$

Step 11.3.3

The left side $0.\overline{90}$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 11.4

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

$x < \frac{2}{3}$ True

$\frac{2}{3} < x < 6$ False

$x > 6$ True

$x < \frac{2}{3}$ True

$\frac{2}{3} < x < 6$ False

$x > 6$ True

Step 12

The solution consists of all of the true intervals.

$x < \frac{2}{3}$ or $x > 6$

Step 13

Convert the inequality to interval notation.

$(- \infty, \frac{2}{3}) \cup (6, \infty)$

Step 14