Precalculus Examples

$\frac{9 x^{2} + 4 y^{2} - 12 x y}{4 y^{2} - 9 x^{2}}$

Step 1

Set the denominator in $\frac{9 x^{2} + 4 y^{2} - 12 x y}{4 y^{2} - 9 x^{2}}$ equal to $0$ to find where the expression is undefined.

$4 y^{2} - 9 x^{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $4 y^{2}$ from both sides of the equation.

$- 9 x^{2} = - 4 y^{2}$

Step 2.2

Divide each term in $- 9 x^{2} = - 4 y^{2}$ by $- 9$ and simplify.

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Step 2.2.1

Divide each term in $- 9 x^{2} = - 4 y^{2}$ by $- 9$ .

$\frac{- 9 x^{2}}{- 9} = \frac{- 4 y^{2}}{- 9}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 9$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 9 x^{2}}{- 9} = \frac{- 4 y^{2}}{- 9}$

Step 2.2.2.1.2

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

$x^{2} = \frac{- 4 y^{2}}{- 9}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Dividing two negative values results in a positive value.

$x^{2} = \frac{4 y^{2}}{9}$

Step 2.3

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

$x = \pm \sqrt{\frac{4 y^{2}}{9}}$

Step 2.4

Simplify $\pm \sqrt{\frac{4 y^{2}}{9}}$ .

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Step 2.4.1

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$x = \pm \sqrt{\frac{{(2 y)}^{2}}{9}}$

Step 2.4.2

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{\frac{{(2 y)}^{2}}{3^{2}}}$

Step 2.4.3

Rewrite $\frac{{(2 y)}^{2}}{3^{2}}$ as ${(\frac{2 y}{3})}^{2}$ .

$x = \pm \sqrt{{(\frac{2 y}{3})}^{2}}$

Step 2.4.4

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

$x = \pm \frac{2 y}{3}$

Step 2.5

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

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Step 2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{2 y}{3}$

Step 2.5.2

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

$x = - \frac{2 y}{3}$

Step 2.5.3

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

$x = \frac{2 y}{3}$

$x = - \frac{2 y}{3}$

$x = \frac{2 y}{3}$

$x = - \frac{2 y}{3}$

$x = \frac{2 y}{3}$

$x = - \frac{2 y}{3}$

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$