Precalculus Examples

$\frac{x^{2} - 1}{5 x y} \cdot \frac{x^{2} y}{1 + x}$

Step 1

Set the denominator in $\frac{x^{2} - 1}{5 x y}$ equal to $0$ to find where the expression is undefined.

$5 x y = 0$

Step 2

Divide each term in $5 x y = 0$ by $5 y$ and simplify.

Tap for more steps...

Step 2.1

Divide each term in $5 x y = 0$ by $5 y$ .

$\frac{5 x y}{5 y} = \frac{0}{5 y}$

Step 2.2

Simplify the left side.

Tap for more steps...

Step 2.2.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 2.2.1.1

Cancel the common factor.

$\frac{5 x y}{5 y} = \frac{0}{5 y}$

Step 2.2.1.2

Rewrite the expression.

$\frac{x y}{y} = \frac{0}{5 y}$

Step 2.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 2.2.2.1

Cancel the common factor.

$\frac{x y}{y} = \frac{0}{5 y}$

Step 2.2.2.2

Divide $x$ by $1$ .

$x = \frac{0}{5 y}$

Step 2.3

Simplify the right side.

Tap for more steps...

Step 2.3.1

Cancel the common factor of $0$ and $5$ .

Tap for more steps...

Step 2.3.1.1

Factor $5$ out of $0$ .

$x = \frac{5 (0)}{5 y}$

Step 2.3.1.2

Cancel the common factors.

Tap for more steps...

Step 2.3.1.2.1

Factor $5$ out of $5 y$ .

$x = \frac{5 (0)}{5 (y)}$

Step 2.3.1.2.2

Cancel the common factor.

$x = \frac{5 \cdot 0}{5 y}$

Step 2.3.1.2.3

Rewrite the expression.

$x = \frac{0}{y}$

Step 2.3.2

Divide $0$ by $y$ .

$x = 0$

Step 3

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

$1 + x = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

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