Precalculus Examples

$\frac{6 x^{2} + 5 x y - 6 y^{2}}{3 x^{2} - 5 x y + 2 y^{2}}$

Step 1

Set the denominator in $\frac{6 x^{2} + 5 x y - 6 y^{2}}{3 x^{2} - 5 x y + 2 y^{2}}$ equal to $0$ to find where the expression is undefined.

$3 x^{2} - 5 x y + 2 y^{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.2

Substitute the values $a = 3$ , $b = - 5 y$ , and $c = 2 y^{2}$ into the quadratic formula and solve for $x$ .

$\frac{5 y \pm \sqrt{{(- 5 y)}^{2} - 4 \cdot (3 \cdot (2 y^{2}))}}{2 \cdot 3}$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the numerator.

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Step 2.3.1.1

Apply the product rule to $- 5 y$ .

$x = \frac{5 y \pm \sqrt{{(- 5)}^{2} y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.3.1.2

Raise $- 5$ to the power of $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.3.1.3

Multiply $- 4 \cdot 3 \cdot 2$ .

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Step 2.3.1.3.1

Multiply $- 4$ by $3$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 12 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.3.1.3.2

Multiply $- 12$ by $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 24 y^{2}}}{2 \cdot 3}$

Step 2.3.1.4

Subtract $24 y^{2}$ from $25 y^{2}$ .

$x = \frac{5 y \pm \sqrt{y^{2}}}{2 \cdot 3}$

Step 2.3.1.5

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

$x = \frac{5 y \pm y}{2 \cdot 3}$

Step 2.3.2

Multiply $2$ by $3$ .

$x = \frac{5 y \pm y}{6}$

Step 2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.4.1

Simplify the numerator.

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Step 2.4.1.1

Apply the product rule to $- 5 y$ .

$x = \frac{5 y \pm \sqrt{{(- 5)}^{2} y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.4.1.2

Raise $- 5$ to the power of $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.4.1.3

Multiply $- 4 \cdot 3 \cdot 2$ .

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Step 2.4.1.3.1

Multiply $- 4$ by $3$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 12 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.4.1.3.2

Multiply $- 12$ by $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 24 y^{2}}}{2 \cdot 3}$

Step 2.4.1.4

Subtract $24 y^{2}$ from $25 y^{2}$ .

$x = \frac{5 y \pm \sqrt{y^{2}}}{2 \cdot 3}$

Step 2.4.1.5

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

$x = \frac{5 y \pm y}{2 \cdot 3}$

Step 2.4.2

Multiply $2$ by $3$ .

$x = \frac{5 y \pm y}{6}$

Step 2.4.3

Change the $\pm$ to $+$ .

$x = \frac{5 y + y}{6}$

Step 2.4.4

Add $5 y$ and $y$ .

$x = \frac{6 y}{6}$

Step 2.4.5

Cancel the common factor of $6$ .

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Step 2.4.5.1

Rewrite.

$x = 1 y$

Step 2.4.5.2

Multiply $y$ by $1$ .

$x = y$

Step 2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

Apply the product rule to $- 5 y$ .

$x = \frac{5 y \pm \sqrt{{(- 5)}^{2} y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.5.1.2

Raise $- 5$ to the power of $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 4 \cdot 3 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.5.1.3

Multiply $- 4 \cdot 3 \cdot 2$ .

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Step 2.5.1.3.1

Multiply $- 4$ by $3$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 12 \cdot (2 y^{2})}}{2 \cdot 3}$

Step 2.5.1.3.2

Multiply $- 12$ by $2$ .

$x = \frac{5 y \pm \sqrt{25 y^{2} - 24 y^{2}}}{2 \cdot 3}$

Step 2.5.1.4

Subtract $24 y^{2}$ from $25 y^{2}$ .

$x = \frac{5 y \pm \sqrt{y^{2}}}{2 \cdot 3}$

Step 2.5.1.5

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

$x = \frac{5 y \pm y}{2 \cdot 3}$

Step 2.5.2

Multiply $2$ by $3$ .

$x = \frac{5 y \pm y}{6}$

Step 2.5.3

Change the $\pm$ to $-$ .

$x = \frac{5 y - y}{6}$

Step 2.5.4

Subtract $y$ from $5 y$ .

$x = \frac{4 y}{6}$

Step 2.5.5

Cancel the common factor of $4$ and $6$ .

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Step 2.5.5.1

Factor $2$ out of $4 y$ .

$x = \frac{2 (2 y)}{6}$

Step 2.5.5.2

Cancel the common factors.

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Step 2.5.5.2.1

Factor $2$ out of $6$ .

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

Step 2.5.5.2.2

Cancel the common factor.

$x = \frac{2 (2 y)}{2 \cdot 3}$

Step 2.5.5.2.3

Rewrite the expression.

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

Step 2.6

The final answer is the combination of both solutions.

$x = y$

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

$x = y$

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

Step 3

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$