Precalculus Examples

$\frac{x^{2} - 9}{y^{2} - 25} \div \frac{2 x^{2} - 6 x}{3 y^{2} - 15 y} + \frac{3 - 1.5 y}{y + 5}$

Step 1

Set the denominator in $\frac{x^{2} - 9}{y^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$y^{2} - 25 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $25$ to both sides of the equation.

$y^{2} = 25$

Step 2.2

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

$y = \pm \sqrt{25}$

Step 2.3

Simplify $\pm \sqrt{25}$ .

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Step 2.3.1

Rewrite $25$ as $5^{2}$ .

$y = \pm \sqrt{5^{2}}$

Step 2.3.2

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

$y = \pm 5$

Step 2.4

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

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Step 2.4.1

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

$y = 5$

Step 2.4.2

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

$y = - 5$

Step 2.4.3

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

$y = 5, - 5$

Step 3

Set the denominator in $\frac{2 x^{2} - 6 x}{3 y^{2} - 15 y}$ equal to $0$ to find where the expression is undefined.

$3 y^{2} - 15 y = 0$

Step 4

Solve for $x$ .

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Step 4.1

Factor $3 y$ out of $3 y^{2} - 15 y$ .

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Step 4.1.1

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

$3 y (y) - 15 y = 0$

Step 4.1.2

Factor $3 y$ out of $- 15 y$ .

$3 y (y) + 3 y (- 5) = 0$

Step 4.1.3

Factor $3 y$ out of $3 y (y) + 3 y (- 5)$ .

$3 y (y - 5) = 0$

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 $0$ .

$y = 0$

$y - 5 = 0$

Step 4.3

Set $y$ equal to $0$ .

$y = 0$

Step 4.4

Set $y - 5$ equal to $0$ and solve for $y$ .

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Step 4.4.1

Set $y - 5$ equal to $0$ .

$y - 5 = 0$

Step 4.4.2

Add $5$ to both sides of the equation.

$y = 5$

Step 4.5

The final solution is all the values that make $3 y (y - 5) = 0$ true.

$y = 0, 5$

Step 5

Set the denominator in $\frac{x^{2} - 9}{y^{2} - 25} \div \frac{2 x^{2} - 6 x}{3 y^{2} - 15 y}$ equal to $0$ to find where the expression is undefined.

$\frac{2 x^{2} - 6 x}{3 y^{2} - 15 y} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$2 x^{2} - 6 x = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Factor $2 x$ out of $2 x^{2} - 6 x$ .

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Step 6.2.1.1

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

$2 x (x) - 6 x = 0$

Step 6.2.1.2

Factor $2 x$ out of $- 6 x$ .

$2 x (x) + 2 x (- 3) = 0$

Step 6.2.1.3

Factor $2 x$ out of $2 x (x) + 2 x (- 3)$ .

$2 x (x - 3) = 0$

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 $0$ .

$x = 0$

$x - 3 = 0$

Step 6.2.3

Set $x$ equal to $0$ .

$x = 0$

Step 6.2.4

Set $x - 3$ equal to $0$ and solve for $x$ .

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Step 6.2.4.1

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Step 6.2.4.2

Add $3$ to both sides of the equation.

$x = 3$

Step 6.2.5

The final solution is all the values that make $2 x (x - 3) = 0$ true.

$x = 0, 3$

Step 7

Set the denominator in $\frac{3 - 1.5 y}{y + 5}$ equal to $0$ to find where the expression is undefined.

$y + 5 = 0$

Step 8

Subtract $5$ from both sides of the equation.

$y = - 5$

Step 9

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

Interval Notation:

$(- \infty, - 5) \cup (- 5, 0) \cup (0, 3) \cup (3, 5) \cup (5, \infty)$

Set-Builder Notation:

${x | x \neq 0, 3, 5, - 5}$