Algebra Examples

$\frac{y^{2} + 15 y + 26}{12 y^{2}} \div \frac{3 y^{3} + 39 y^{2}}{y^{2} - 9 y}$

Step 1

Set the denominator in $\frac{y^{2} + 15 y + 26}{12 y^{2}}$ equal to $0$ to find where the expression is undefined.

$12 y^{2} = 0$

Step 2

Solve for $y$ .

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Step 2.1

Divide each term in $12 y^{2} = 0$ by $12$ and simplify.

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Step 2.1.1

Divide each term in $12 y^{2} = 0$ by $12$ .

$\frac{12 y^{2}}{12} = \frac{0}{12}$

Step 2.1.2

Simplify the left side.

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Step 2.1.2.1

Cancel the common factor of $12$ .

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Step 2.1.2.1.1

Cancel the common factor.

$\frac{12 y^{2}}{12} = \frac{0}{12}$

Step 2.1.2.1.2

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

$y^{2} = \frac{0}{12}$

Step 2.1.3

Simplify the right side.

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Step 2.1.3.1

Divide $0$ by $12$ .

$y^{2} = 0$

Step 2.2

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

$y = \pm \sqrt{0}$

Step 2.3

Simplify $\pm \sqrt{0}$ .

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Step 2.3.1

Rewrite $0$ as $0^{2}$ .

$y = \pm \sqrt{0^{2}}$

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 $\frac{3 y^{3} + 39 y^{2}}{y^{2} - 9 y}$ equal to $0$ to find where the expression is undefined.

$y^{2} - 9 y = 0$

Step 4

Solve for $y$ .

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Step 4.1

Factor $y$ out of $y^{2} - 9 y$ .

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Step 4.1.1

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

$y \cdot y - 9 y = 0$

Step 4.1.2

Factor $y$ out of $- 9 y$ .

$y \cdot y + y \cdot - 9 = 0$

Step 4.1.3

Factor $y$ out of $y \cdot y + y \cdot - 9$ .

$y (y - 9) = 0$

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 $0$ .

$y = 0$

$y - 9 = 0$

Step 4.3

Set $y$ equal to $0$ .

$y = 0$

Step 4.4

Set $y - 9$ equal to $0$ and solve for $y$ .

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Step 4.4.1

Set $y - 9$ equal to $0$ .

$y - 9 = 0$

Step 4.4.2

Add $9$ to both sides of the equation.

$y = 9$

Step 4.5

The final solution is all the values that make $y (y - 9) = 0$ true.

$y = 0, 9$

Step 5

Set the denominator in $\frac{y^{2} + 15 y + 26}{12 y^{2}} \div \frac{3 y^{3} + 39 y^{2}}{y^{2} - 9 y}$ equal to $0$ to find where the expression is undefined.

$\frac{3 y^{3} + 39 y^{2}}{y^{2} - 9 y} = 0$

Step 6

Solve for $y$ .

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Step 6.1

Set the numerator equal to zero.

$3 y^{3} + 39 y^{2} = 0$

Step 6.2

Solve the equation for $y$ .

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Step 6.2.1

Factor $3 y^{2}$ out of $3 y^{3} + 39 y^{2}$ .

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Step 6.2.1.1

Factor $3 y^{2}$ out of $3 y^{3}$ .

$3 y^{2} (y) + 39 y^{2} = 0$

Step 6.2.1.2

Factor $3 y^{2}$ out of $39 y^{2}$ .

$3 y^{2} (y) + 3 y^{2} (13) = 0$

Step 6.2.1.3

Factor $3 y^{2}$ out of $3 y^{2} (y) + 3 y^{2} (13)$ .

$3 y^{2} (y + 13) = 0$

Step 6.2.2

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

$y^{2} = 0$

$y + 13 = 0$

Step 6.2.3

Set $y^{2}$ equal to $0$ and solve for $y$ .

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Step 6.2.3.1

Set $y^{2}$ equal to $0$ .

$y^{2} = 0$

Step 6.2.3.2

Solve $y^{2} = 0$ for $y$ .

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Step 6.2.3.2.1

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

$y = \pm \sqrt{0}$

Step 6.2.3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 6.2.3.2.2.1

Rewrite $0$ as $0^{2}$ .

$y = \pm \sqrt{0^{2}}$

Step 6.2.3.2.2.2

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

$y = \pm 0$

Step 6.2.3.2.2.3

Plus or minus $0$ is $0$ .

$y = 0$

Step 6.2.4

Set $y + 13$ equal to $0$ and solve for $y$ .

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Step 6.2.4.1

Set $y + 13$ equal to $0$ .

$y + 13 = 0$

Step 6.2.4.2

Subtract $13$ from both sides of the equation.

$y = - 13$

Step 6.2.5

The final solution is all the values that make $3 y^{2} (y + 13) = 0$ true.

$y = 0, - 13$

Step 6.3

Exclude the solutions that do not make $\frac{3 y^{3} + 39 y^{2}}{y^{2} - 9 y} = 0$ true.

$y = - 13$

Step 7

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$y = - 13, y = 0, y = 9$

Step 8