Algebra Examples

$\frac{x + 2}{x^{2} - 25} + 1 = \frac{8 x}{2 x - 10}$

Step 1

Factor each term.

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Step 1.1

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

$\frac{x + 2}{x^{2} - 5^{2}} + 1 = \frac{8 x}{2 x - 10}$

Step 1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{8 x}{2 x - 10}$

Step 1.3

Factor $2$ out of $2 x - 10$ .

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Step 1.3.1

Factor $2$ out of $2 x$ .

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{8 x}{2 (x) - 10}$

Step 1.3.2

Factor $2$ out of $- 10$ .

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{8 x}{2 x + 2 \cdot - 5}$

Step 1.3.3

Factor $2$ out of $2 x + 2 \cdot - 5$ .

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{8 x}{2 (x - 5)}$

Step 1.4

Reduce the expression $\frac{8 x}{2 (x - 5)}$ by cancelling the common factors.

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Step 1.4.1

Factor $2$ out of $8 x$ .

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{2 (4 x)}{2 (x - 5)}$

Step 1.4.2

Cancel the common factor.

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{2 (4 x)}{2 (x - 5)}$

Step 1.4.3

Rewrite the expression.

$\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{4 x}{x - 5}$

Step 2

Find the LCD of the terms in the equation.

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Step 2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$(x + 5) (x - 5), 1, x - 5$

Step 2.2

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.3

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.4

The LCM of $1, 1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

Step 2.5

The factor for $x + 5$ is $x + 5$ itself.

$(x + 5) = x + 5$

$(x + 5)$ occurs $1$ time.

Step 2.6

The factor for $x - 5$ is $x - 5$ itself.

$(x - 5) = x - 5$

$(x - 5)$ occurs $1$ time.

Step 2.7

The LCM of $x + 5, x - 5, x - 5$ is the result of multiplying all factors the greatest number of times they occur in either term.

$(x + 5) (x - 5)$

Step 3

Multiply each term in $\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{4 x}{x - 5}$ by $(x + 5) (x - 5)$ to eliminate the fractions.

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Step 3.1

Multiply each term in $\frac{x + 2}{(x + 5) (x - 5)} + 1 = \frac{4 x}{x - 5}$ by $(x + 5) (x - 5)$ .

$\frac{x + 2}{(x + 5) (x - 5)} ((x + 5) (x - 5)) + 1 ((x + 5) (x - 5)) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Cancel the common factor of $(x + 5) (x - 5)$ .

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Step 3.2.1.1.1

Cancel the common factor.

$\frac{x + 2}{(x + 5) (x - 5)} ((x + 5) (x - 5)) + 1 ((x + 5) (x - 5)) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.1.2

Rewrite the expression.

$x + 2 + 1 ((x + 5) (x - 5)) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.2

Multiply $(x + 5) (x - 5)$ by $1$ .

$x + 2 + (x + 5) (x - 5) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.3

Expand $(x + 5) (x - 5)$ using the FOIL Method.

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Step 3.2.1.3.1

Apply the distributive property.

$x + 2 + x (x - 5) + 5 (x - 5) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.3.2

Apply the distributive property.

$x + 2 + x \cdot x + x \cdot - 5 + 5 (x - 5) = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.3.3

Apply the distributive property.

$x + 2 + x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.4

Combine the opposite terms in $x \cdot x + x \cdot - 5 + 5 x + 5 \cdot - 5$ .

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Step 3.2.1.4.1

Reorder the factors in the terms $x \cdot - 5$ and $5 x$ .

$x + 2 + x \cdot x - 5 x + 5 x + 5 \cdot - 5 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.4.2

Add $- 5 x$ and $5 x$ .

$x + 2 + x \cdot x + 0 + 5 \cdot - 5 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.4.3

Add $x \cdot x$ and $0$ .

$x + 2 + x \cdot x + 5 \cdot - 5 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.5

Simplify each term.

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Step 3.2.1.5.1

Multiply $x$ by $x$ .

$x + 2 + x^{2} + 5 \cdot - 5 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.1.5.2

Multiply $5$ by $- 5$ .

$x + 2 + x^{2} - 25 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.2.2

Subtract $25$ from $2$ .

$x + x^{2} - 23 = \frac{4 x}{x - 5} ((x + 5) (x - 5))$

Step 3.3

Simplify the right side.

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Step 3.3.1

Cancel the common factor of $x - 5$ .

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Step 3.3.1.1

Factor $x - 5$ out of $(x + 5) (x - 5)$ .

$x + x^{2} - 23 = \frac{4 x}{x - 5} ((x - 5) (x + 5))$

Step 3.3.1.2

Cancel the common factor.

$x + x^{2} - 23 = \frac{4 x}{x - 5} ((x - 5) (x + 5))$

Step 3.3.1.3

Rewrite the expression.

$x + x^{2} - 23 = 4 x (x + 5)$

Step 3.3.2

Apply the distributive property.

$x + x^{2} - 23 = 4 x \cdot x + 4 x \cdot 5$

Step 3.3.3

Multiply $x$ by $x$ by adding the exponents.

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Step 3.3.3.1

Move $x$ .

$x + x^{2} - 23 = 4 (x \cdot x) + 4 x \cdot 5$

Step 3.3.3.2

Multiply $x$ by $x$ .

$x + x^{2} - 23 = 4 x^{2} + 4 x \cdot 5$

Step 3.3.4

Multiply $5$ by $4$ .

$x + x^{2} - 23 = 4 x^{2} + 20 x$

Step 4

Solve the equation.

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Step 4.1

Move all terms containing $x$ to the left side of the equation.

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Step 4.1.1

Subtract $4 x^{2}$ from both sides of the equation.

$x + x^{2} - 23 - 4 x^{2} = 20 x$

Step 4.1.2

Subtract $20 x$ from both sides of the equation.

$x + x^{2} - 23 - 4 x^{2} - 20 x = 0$

Step 4.1.3

Subtract $20 x$ from $x$ .

$x^{2} - 23 - 4 x^{2} - 19 x = 0$

Step 4.1.4

Subtract $4 x^{2}$ from $x^{2}$ .

$- 3 x^{2} - 23 - 19 x = 0$

Step 4.2

Use the quadratic formula to find the solutions.

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

Step 4.3

Substitute the values $a = - 3$ , $b = - 19$ , and $c = - 23$ into the quadratic formula and solve for $x$ .

$\frac{19 \pm \sqrt{{(- 19)}^{2} - 4 \cdot (- 3 \cdot - 23)}}{2 \cdot - 3}$

Step 4.4

Simplify.

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Step 4.4.1

Simplify the numerator.

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Step 4.4.1.1

Raise $- 19$ to the power of $2$ .

$x = \frac{19 \pm \sqrt{361 - 4 \cdot - 3 \cdot - 23}}{2 \cdot - 3}$

Step 4.4.1.2

Multiply $- 4 \cdot - 3 \cdot - 23$ .

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Step 4.4.1.2.1

Multiply $- 4$ by $- 3$ .

$x = \frac{19 \pm \sqrt{361 + 12 \cdot - 23}}{2 \cdot - 3}$

Step 4.4.1.2.2

Multiply $12$ by $- 23$ .

$x = \frac{19 \pm \sqrt{361 - 276}}{2 \cdot - 3}$

Step 4.4.1.3

Subtract $276$ from $361$ .

$x = \frac{19 \pm \sqrt{85}}{2 \cdot - 3}$

Step 4.4.2

Multiply $2$ by $- 3$ .

$x = \frac{19 \pm \sqrt{85}}{- 6}$

Step 4.4.3

Move the negative in front of the fraction.

$x = - \frac{19 \pm \sqrt{85}}{6}$

Step 4.5

The final answer is the combination of both solutions.

$x = - \frac{19 + \sqrt{85}}{6}, - \frac{19 - \sqrt{85}}{6}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{19 + \sqrt{85}}{6}, - \frac{19 - \sqrt{85}}{6}$

Decimal Form:

$x = - 4.70325740 \dots, - 1.63007592 \dots$