Precalculus Examples

$\frac{25 x^{2} - 4}{x^{2} - 9} \div \frac{5 x - 2}{x + 3}$

Step 1

Set the denominator in $\frac{25 x^{2} - 4}{x^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 9 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 2.2

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

$x = \pm \sqrt{9}$

Step 2.3

Simplify $\pm \sqrt{9}$ .

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Step 2.3.1

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

$x = \pm \sqrt{3^{2}}$

Step 2.3.2

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

$x = \pm 3$

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.

$x = 3$

Step 2.4.2

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

$x = - 3$

Step 2.4.3

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

$x = 3, - 3$

Step 3

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

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

Set the denominator in $\frac{25 x^{2} - 4}{x^{2} - 9} \div \frac{5 x - 2}{x + 3}$ equal to $0$ to find where the expression is undefined.

$\frac{5 x - 2}{x + 3} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$5 x - 2 = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Add $2$ to both sides of the equation.

$5 x = 2$

Step 6.2.2

Divide each term in $5 x = 2$ by $5$ and simplify.

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Step 6.2.2.1

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $5$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{5}$

Step 7

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

Interval Notation:

$(- \infty, - 3) \cup (- 3, \frac{2}{5}) \cup (\frac{2}{5}, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq \frac{2}{5}, 3, - 3}$

Step 8