Precalculus Examples

$\frac{a + 3}{a^{2} - 1} - \frac{1}{a^{2} + a}$

Step 1

Set the denominator in $\frac{a + 3}{a^{2} - 1}$ equal to $0$ to find where the expression is undefined.

$a^{2} - 1 = 0$

Step 2

Solve for $a$ .

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Step 2.1

Add $1$ to both sides of the equation.

$a^{2} = 1$

Step 2.2

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

$a = \pm \sqrt{1}$

Step 2.3

Any root of $1$ is $1$ .

$a = \pm 1$

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.

$a = 1$

Step 2.4.2

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

$a = - 1$

Step 2.4.3

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

$a = 1, - 1$

Step 3

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

$a^{2} + a = 0$

Step 4

Solve for $a$ .

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Step 4.1

Factor $a$ out of $a^{2} + a$ .

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Step 4.1.1

Factor $a$ out of $a^{2}$ .

$a \cdot a + a = 0$

Step 4.1.2

Raise $a$ to the power of $1$ .

$a \cdot a + a = 0$

Step 4.1.3

Factor $a$ out of $a^{1}$ .

$a \cdot a + a \cdot 1 = 0$

Step 4.1.4

Factor $a$ out of $a \cdot a + a \cdot 1$ .

$a (a + 1) = 0$

Step 4.2

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

$a = 0$

$a + 1 = 0$

Step 4.3

Set $a$ equal to $0$ .

$a = 0$

Step 4.4

Set $a + 1$ equal to $0$ and solve for $a$ .

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Step 4.4.1

Set $a + 1$ equal to $0$ .

$a + 1 = 0$

Step 4.4.2

Subtract $1$ from both sides of the equation.

$a = - 1$

Step 4.5

The final solution is all the values that make $a (a + 1) = 0$ true.

$a = 0, - 1$

Step 5

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 0) \cup (0, 1) \cup (1, \infty)$

Set-Builder Notation:

${a | a \neq 0, 1, - 1}$

Step 6