Find the Domain 1/(x^2-81)-4/(x-9)=1/(x+9)

$\frac{1}{x^{2} - 81} - \frac{4}{x - 9} = \frac{1}{x + 9}$

Step 1

Set the denominator in $\frac{1}{x^{2} - 81}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 81 = 0$

Step 2

Solve for $x$ .

Add $81$ to both sides of the equation.

$x^{2} = 81$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{81}$

Simplify $\pm \sqrt{81}$ .

Rewrite $81$ as $9^{2}$ .

$x = \pm \sqrt{9^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 9$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 9$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 9$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 9, - 9$

Step 3

Set the denominator in $\frac{4}{x - 9}$ equal to $0$ to find where the expression is undefined.

$x - 9 = 0$

Step 4

Add $9$ to both sides of the equation.

$x = 9$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 9, - 9}$

Step 6