Algebra Examples

$\frac{x^{2} + 10 x + 25}{x - 5} \div \frac{x^{2} - 25}{5 x + 10}$

Step 1

Set the denominator in $\frac{x^{2} + 10 x + 25}{x - 5}$ equal to $0$ to find where the expression is undefined.

$x - 5 = 0$

Step 2

Add $5$ to both sides of the equation.

$x = 5$

Step 3

Set the denominator in $\frac{x^{2} - 25}{5 x + 10}$ equal to $0$ to find where the expression is undefined.

$5 x + 10 = 0$

Step 4

Solve for $x$ .

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Step 4.1

Subtract $10$ from both sides of the equation.

$5 x = - 10$

Step 4.2

Divide each term in $5 x = - 10$ by $5$ and simplify.

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Step 4.2.1

Divide each term in $5 x = - 10$ by $5$ .

$\frac{5 x}{5} = \frac{- 10}{5}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $5$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 10}{5}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 10}{5}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Divide $- 10$ by $5$ .

$x = - 2$

Step 5

Set the denominator in $\frac{x^{2} + 10 x + 25}{x - 5} \div \frac{x^{2} - 25}{5 x + 10}$ equal to $0$ to find where the expression is undefined.

$\frac{x^{2} - 25}{5 x + 10} = 0$

Step 6

Solve for $x$ .

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Step 6.1

Set the numerator equal to zero.

$x^{2} - 25 = 0$

Step 6.2

Solve the equation for $x$ .

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Step 6.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 6.2.2

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

$x = \pm \sqrt{25}$

Step 6.2.3

Simplify $\pm \sqrt{25}$ .

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Step 6.2.3.1

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

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

Step 6.2.3.2

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

$x = \pm 5$

Step 6.2.4

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

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Step 6.2.4.1

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

$x = 5$

Step 6.2.4.2

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

$x = - 5$

Step 6.2.4.3

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

$x = 5, - 5$

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 $0$ .

$x = - 5, x = - 2, x = 5$

Step 8