Precalculus Examples

$\frac{a}{a^{2} - 9} + \frac{3}{9 - a^{2}}$

Step 1

Set the denominator in $\frac{a}{a^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$a^{2} - 9 = 0$

Step 2

Solve for $a$ .

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Step 2.1

Add $9$ to both sides of the equation.

$a^{2} = 9$

Step 2.2

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

$a = \pm \sqrt{9}$

Step 2.3

Simplify $\pm \sqrt{9}$ .

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Step 2.3.1

Rewrite $9$ as $3^{2}$ .

$a = \pm \sqrt{3^{2}}$

Step 2.3.2

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

$a = \pm 3$

Step 2.4

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

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Step 2.4.1

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

$a = 3$

Step 2.4.2

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

$a = - 3$

Step 2.4.3

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

$a = 3, - 3$

Step 3

Set the denominator in $\frac{3}{9 - a^{2}}$ equal to $0$ to find where the expression is undefined.

$9 - a^{2} = 0$

Step 4

Solve for $a$ .

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Step 4.1

Subtract $9$ from both sides of the equation.

$- a^{2} = - 9$

Step 4.2

Divide each term in $- a^{2} = - 9$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- a^{2} = - 9$ by $- 1$ .

$\frac{- a^{2}}{- 1} = \frac{- 9}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{a^{2}}{1} = \frac{- 9}{- 1}$

Step 4.2.2.2

Divide $a^{2}$ by $1$ .

$a^{2} = \frac{- 9}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 9$ by $- 1$ .

$a^{2} = 9$

Step 4.3

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

$a = \pm \sqrt{9}$

Step 4.4

Simplify $\pm \sqrt{9}$ .

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Step 4.4.1

Rewrite $9$ as $3^{2}$ .

$a = \pm \sqrt{3^{2}}$

Step 4.4.2

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

$a = \pm 3$

Step 4.5

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

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Step 4.5.1

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

$a = 3$

Step 4.5.2

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

$a = - 3$

Step 4.5.3

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

$a = 3, - 3$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${a | a \neq 3, - 3}$

Step 6