Precalculus Examples

$\frac{48 x^{5}}{49 y^{4}} \cdot \frac{7 y^{2}}{16 x^{3}}$

Step 1

Set the denominator in $\frac{48 x^{5}}{49 y^{4}}$ equal to $0$ to find where the expression is undefined.

$49 y^{4} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Divide each term in $49 y^{4} = 0$ by $49$ and simplify.

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Step 2.1.1

Divide each term in $49 y^{4} = 0$ by $49$ .

$\frac{49 y^{4}}{49} = \frac{0}{49}$

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Cancel the common factor of $49$ .

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Step 2.1.2.1.1

Cancel the common factor.

$\frac{49 y^{4}}{49} = \frac{0}{49}$

Step 2.1.2.1.2

Divide $y^{4}$ by $1$ .

$y^{4} = \frac{0}{49}$

Step 2.1.3

Simplify the right side.

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Step 2.1.3.1

Divide $0$ by $49$ .

$y^{4} = 0$

Step 2.2

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

$y = \pm \sqrt[4]{0}$

Step 2.3

Simplify $\pm \sqrt[4]{0}$ .

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Step 2.3.1

Rewrite $0$ as $0^{4}$ .

$y = \pm \sqrt[4]{0^{4}}$

Step 2.3.2

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

$y = \pm 0$

Step 2.3.3

Plus or minus $0$ is $0$ .

$y = 0$

Step 3

Set the denominator in $\frac{7 y^{2}}{16 x^{3}}$ equal to $0$ to find where the expression is undefined.

$16 x^{3} = 0$

Step 4

Solve for $x$ .

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Step 4.1

Divide each term in $16 x^{3} = 0$ by $16$ and simplify.

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Step 4.1.1

Divide each term in $16 x^{3} = 0$ by $16$ .

$\frac{16 x^{3}}{16} = \frac{0}{16}$

Step 4.1.2

Simplify the left side.

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Step 4.1.2.1

Cancel the common factor of $16$ .

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Step 4.1.2.1.1

Cancel the common factor.

$\frac{16 x^{3}}{16} = \frac{0}{16}$

Step 4.1.2.1.2

Divide $x^{3}$ by $1$ .

$x^{3} = \frac{0}{16}$

Step 4.1.3

Simplify the right side.

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Step 4.1.3.1

Divide $0$ by $16$ .

$x^{3} = 0$

Step 4.2

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

$x = \sqrt[3]{0}$

Step 4.3

Simplify $\sqrt[3]{0}$ .

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Step 4.3.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 4.3.2

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

$x = 0$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${x | x \neq 0}$