Precalculus Examples

$\frac{4 a b}{c x + d x} \cdot \frac{a x + b x}{2 a b}$

Step 1

Set the denominator in $\frac{4 a b}{c x + d x}$ equal to $0$ to find where the expression is undefined.

$c x + d x = 0$

Step 2

Solve for $a$ .

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Step 2.1

Subtract $d x$ from both sides of the equation.

$c x = - d x$

Step 2.2

Divide each term in $c x = - d x$ by $x$ and simplify.

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Step 2.2.1

Divide each term in $c x = - d x$ by $x$ .

$\frac{c x}{x} = \frac{- d x}{x}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $x$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{c x}{x} = \frac{- d x}{x}$

Step 2.2.2.1.2

Divide $c$ by $1$ .

$c = \frac{- d x}{x}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Cancel the common factor of $x$ .

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Step 2.2.3.1.1

Cancel the common factor.

$c = \frac{- d x}{x}$

Step 2.2.3.1.2

Divide $- d$ by $1$ .

$c = - d$

Step 3

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

$2 a b = 0$

Step 4

Divide each term in $2 a b = 0$ by $2 b$ and simplify.

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Step 4.1

Divide each term in $2 a b = 0$ by $2 b$ .

$\frac{2 a b}{2 b} = \frac{0}{2 b}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $2$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{2 a b}{2 b} = \frac{0}{2 b}$

Step 4.2.1.2

Rewrite the expression.

$\frac{a b}{b} = \frac{0}{2 b}$

Step 4.2.2

Cancel the common factor of $b$ .

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Step 4.2.2.1

Cancel the common factor.

$\frac{a b}{b} = \frac{0}{2 b}$

Step 4.2.2.2

Divide $a$ by $1$ .

$a = \frac{0}{2 b}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Cancel the common factor of $0$ and $2$ .

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Step 4.3.1.1

Factor $2$ out of $0$ .

$a = \frac{2 (0)}{2 b}$

Step 4.3.1.2

Cancel the common factors.

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Step 4.3.1.2.1

Factor $2$ out of $2 b$ .

$a = \frac{2 (0)}{2 (b)}$

Step 4.3.1.2.2

Cancel the common factor.

$a = \frac{2 \cdot 0}{2 b}$

Step 4.3.1.2.3

Rewrite the expression.

$a = \frac{0}{b}$

Step 4.3.2

Divide $0$ by $b$ .

$a = 0$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${a | a \neq 0}$