Solve the Inequality for x (x-6)/((x-3)(x-9))<0

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

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

$x - 6 = 0$

$x - 3 = 0$

$x - 9 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Add $3$ to both sides of the equation.

$x = 3$

Add $9$ to both sides of the equation.

$x = 9$

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

$x = 6$

$x = 3$

$x = 9$

Consolidate the solutions.

$x = 6, 3, 9$

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

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

$(x - 3) (x - 9) = 0$

Solve for $x$ .

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

$x - 3 = 0$

$x - 9 = 0$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

The final solution is all the values that make $(x - 3) (x - 9) = 0$ true.

$x = 3, 9$

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

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

Use each root to create test intervals.

$x < 3$

$3 < x < 6$

$6 < x < 9$

$x > 9$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = 0$

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

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

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

True

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

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

$x = 4$

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

$\frac{(4) - 6}{((4) - 3) ((4) - 9)} < 0$

The left side $0.4$ is not less than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 8$

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

$\frac{(8) - 6}{((8) - 3) ((8) - 9)} < 0$

The left side $- 0.4$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 12$

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

$\frac{(12) - 6}{((12) - 3) ((12) - 9)} < 0$

The left side $0.\overline{2}$ is not less than the right side $0$ , which means that the given statement is false.

False

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

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

$x < 3 or 6 < x < 9$

Interval Notation:

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

Solve the Inequality for x (x-6)/((x-3)(x-9))<0

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