Precalculus Examples

$\frac{- 1}{x - 6} < \frac{2}{9 - x}$

Step 1

Subtract $\frac{2}{9 - x}$ from both sides of the inequality.

$\frac{- 1}{x - 6} - \frac{2}{9 - x} < 0$

Step 2

Simplify $\frac{- 1}{x - 6} - \frac{2}{9 - x}$ .

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Step 2.1

Move the negative in front of the fraction.

$- \frac{1}{x - 6} - \frac{2}{9 - x} < 0$

Step 2.2

To write $- \frac{1}{x - 6}$ as a fraction with a common denominator, multiply by $\frac{9 - x}{9 - x}$ .

$- \frac{1}{x - 6} \cdot \frac{9 - x}{9 - x} - \frac{2}{9 - x} < 0$

Step 2.3

To write $- \frac{2}{9 - x}$ as a fraction with a common denominator, multiply by $\frac{x - 6}{x - 6}$ .

$- \frac{1}{x - 6} \cdot \frac{9 - x}{9 - x} - \frac{2}{9 - x} \cdot \frac{x - 6}{x - 6} < 0$

Step 2.4

Write each expression with a common denominator of $(x - 6) (9 - x)$ , by multiplying each by an appropriate factor of $1$ .

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Step 2.4.1

Multiply $\frac{1}{x - 6}$ by $\frac{9 - x}{9 - x}$ .

$- \frac{9 - x}{(x - 6) (9 - x)} - \frac{2}{9 - x} \cdot \frac{x - 6}{x - 6} < 0$

Step 2.4.2

Multiply $\frac{2}{9 - x}$ by $\frac{x - 6}{x - 6}$ .

$- \frac{9 - x}{(x - 6) (9 - x)} - \frac{2 (x - 6)}{(9 - x) (x - 6)} < 0$

Step 2.4.3

Reorder the factors of $(x - 6) (9 - x)$ .

$- \frac{9 - x}{(9 - x) (x - 6)} - \frac{2 (x - 6)}{(9 - x) (x - 6)} < 0$

Step 2.5

Combine the numerators over the common denominator.

$\frac{- (9 - x) - 2 (x - 6)}{(9 - x) (x - 6)} < 0$

Step 2.6

Simplify the numerator.

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Step 2.6.1

Factor $- 1$ out of $- (9 - x) - 2 (x - 6)$ .

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Step 2.6.1.1

Reorder $9$ and $- x$ .

$\frac{- (- x + 9) - 2 (x - 6)}{(9 - x) (x - 6)} < 0$

Step 2.6.1.2

Factor $- 1$ out of $- 2 (x - 6)$ .

$\frac{- (- x + 9) - (2 (x - 6))}{(9 - x) (x - 6)} < 0$

Step 2.6.1.3

Factor $- 1$ out of $- (- x + 9) - (2 (x - 6))$ .

$\frac{- (- x + 9 + 2 (x - 6))}{(9 - x) (x - 6)} < 0$

Step 2.6.2

Apply the distributive property.

$\frac{- (- x + 9 + 2 x + 2 \cdot - 6)}{(9 - x) (x - 6)} < 0$

Step 2.6.3

Multiply $2$ by $- 6$ .

$\frac{- (- x + 9 + 2 x - 12)}{(9 - x) (x - 6)} < 0$

Step 2.6.4

Add $- x$ and $2 x$ .

$\frac{- (x + 9 - 12)}{(9 - x) (x - 6)} < 0$

Step 2.6.5

Subtract $12$ from $9$ .

$\frac{- (x - 3)}{(9 - x) (x - 6)} < 0$

Step 2.7

Move the negative in front of the fraction.

$- \frac{x - 3}{(9 - x) (x - 6)} < 0$

Step 3

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

$x - 3 = 0$

$9 - x = 0$

$x - 6 = 0$

Step 4

Add $3$ to both sides of the equation.

$x = 3$

Step 5

Subtract $9$ from both sides of the equation.

$- x = - 9$

Step 6

Divide each term in $- x = - 9$ by $- 1$ and simplify.

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Step 6.1

Divide each term in $- x = - 9$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 9}{- 1}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 9}{- 1}$

Step 6.2.2

Divide $x$ by $1$ .

$x = \frac{- 9}{- 1}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Divide $- 9$ by $- 1$ .

$x = 9$

Step 7

Add $6$ to both sides of the equation.

$x = 6$

Step 8

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

$x = 3$

$x = 9$

$x = 6$

Step 9

Consolidate the solutions.

$x = 3, 9, 6$

Step 10

Find the domain of $- \frac{x - 3}{(9 - x) (x - 6)}$ .

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Step 10.1

Set the denominator in $\frac{x - 3}{(9 - x) (x - 6)}$ equal to $0$ to find where the expression is undefined.

$(9 - x) (x - 6) = 0$

Step 10.2

Solve for $x$ .

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Step 10.2.1

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

$9 - x = 0$

$x - 6 = 0$

Step 10.2.2

Set $9 - x$ equal to $0$ and solve for $x$ .

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Step 10.2.2.1

Set $9 - x$ equal to $0$ .

$9 - x = 0$

Step 10.2.2.2

Solve $9 - x = 0$ for $x$ .

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Step 10.2.2.2.1

Subtract $9$ from both sides of the equation.

$- x = - 9$

Step 10.2.2.2.2

Divide each term in $- x = - 9$ by $- 1$ and simplify.

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Step 10.2.2.2.2.1

Divide each term in $- x = - 9$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 9}{- 1}$

Step 10.2.2.2.2.2

Simplify the left side.

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Step 10.2.2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 9}{- 1}$

Step 10.2.2.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 9}{- 1}$

Step 10.2.2.2.2.3

Simplify the right side.

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Step 10.2.2.2.2.3.1

Divide $- 9$ by $- 1$ .

$x = 9$

Step 10.2.3

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 10.2.3.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 10.2.3.2

Add $6$ to both sides of the equation.

$x = 6$

Step 10.2.4

The final solution is all the values that make $(9 - x) (x - 6) = 0$ true.

$x = 9, 6$

Step 10.3

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

$(- \infty, 6) \cup (6, 9) \cup (9, \infty)$

Step 11

Use each root to create test intervals.

$x < 3$

$3 < x < 6$

$6 < x < 9$

$x > 9$

Step 12

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 12.1

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

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Step 12.1.1

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

$x = 0$

Step 12.1.2

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

$\frac{- 1}{(0) - 6} < \frac{2}{9 - (0)}$

Step 12.1.3

The left side $0.1\overline{6}$ is less than the right side $0.\overline{2}$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $3 < x < 6$ to see if it makes the inequality true.

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Step 12.2.1

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

$x = 4$

Step 12.2.2

Replace $x$ with $4$ in the original inequality.

$\frac{- 1}{(4) - 6} < \frac{2}{9 - (4)}$

Step 12.2.3

The left side $0.5$ is not less than the right side $0.4$ , which means that the given statement is false.

False

Step 12.3

Test a value on the interval $6 < x < 9$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $6 < x < 9$ and see if this value makes the original inequality true.

$x = 8$

Step 12.3.2

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

$\frac{- 1}{(8) - 6} < \frac{2}{9 - (8)}$

Step 12.3.3

The left side $- 0.5$ is less than the right side $2$ , which means that the given statement is always true.

True

Step 12.4

Test a value on the interval $x > 9$ to see if it makes the inequality true.

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Step 12.4.1

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Step 12.4.2

Replace $x$ with $12$ in the original inequality.

$\frac{- 1}{(12) - 6} < \frac{2}{9 - (12)}$

Step 12.4.3

The left side $- 0.1\overline{6}$ is not less than the right side $- 0.\overline{6}$ , which means that the given statement is false.

False

Step 12.5

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

$x < 3$ True

$3 < x < 6$ False

$6 < x < 9$ True

$x > 9$ False

$x < 3$ True

$3 < x < 6$ False

$6 < x < 9$ True

$x > 9$ False

Step 13

The solution consists of all of the true intervals.

$x < 3$ or $6 < x < 9$

Step 14

Convert the inequality to interval notation.

$(- \infty, 3) \cup (6, 9)$

Step 15