Linear Algebra Examples

\frac{1}{x + 5} + \frac{10}{x^{2} - 25} = 1

$\frac{1}{x + 5} + \frac{10}{x^{2} - 25} = 1$

Step 1

Set the denominator in

\frac{1}{x + 5}

$\frac{1}{x + 5}$ equal to

0

$0$ to find where the expression is undefined.

x + 5 = 0

$x + 5 = 0$

Step 2

Subtract

5

$5$ from both sides of the equation.

x = - 5

$x = - 5$

Step 3

Set the denominator in

\frac{10}{x^{2} - 25}

$\frac{10}{x^{2} - 25}$ equal to

0

$0$ to find where the expression is undefined.

x^{2} - 25 = 0

$x^{2} - 25 = 0$

Step 4

Solve for

x

$x$ .

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Step 4.1

Add

25

$25$ to both sides of the equation.

x^{2} = 25

$x^{2} = 25$

Step 4.2

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

x = \pm \sqrt{25}

$x = \pm \sqrt{25}$

Step 4.3

Simplify

\pm \sqrt{25}

$\pm \sqrt{25}$ .

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Step 4.3.1

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

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

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

Step 4.3.2

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

x = \pm 5

$x = \pm 5$

x = \pm 5

$x = \pm 5$

Step 4.4

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

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Step 4.4.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = 5

$x = 5$

Step 4.4.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - 5

$x = - 5$

Step 4.4.3

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

x = 5, - 5

$x = 5, - 5$

x = 5, - 5

$x = 5, - 5$

x = 5, - 5

$x = 5, - 5$

Step 5

The domain is all values of

x

$x$ that make the expression defined.

Interval Notation:

(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)

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

Set-Builder Notation:

{x | x \neq 5, - 5}

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

Step 6

\frac{1}{x + 5} + \frac{10}{x^{2} - 25} = 1

$\frac{1}{x + 5} + \frac{10}{x^{2} - 25} = 1$

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