Solve by Factoring x/(x-2)+1/(x-4)=2/(x^2-6x+8)

$\frac{x}{x - 2} + \frac{1}{x - 4} = \frac{2}{x^{2} - 6 x + 8}$

Move $\frac{2}{x^{2} - 6 x + 8}$ to the left side of the equation by subtracting it from both sides.

$\frac{x}{x - 2} + \frac{1}{x - 4} - \frac{2}{x^{2} - 6 x + 8} = 0$

Simplify $\frac{x}{x - 2} + \frac{1}{x - 4} - \frac{2}{x^{2} - 6 x + 8}$ .

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Factor $x^{2} - 6 x + 8$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$\frac{x}{x - 2} + \frac{1}{x - 4} - \frac{2}{(x - 4) (x - 2)} = 0$

To write $\frac{x}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{x}{x - 2} \cdot \frac{x - 4}{x - 4} + \frac{1}{x - 4} - \frac{2}{(x - 4) (x - 2)} = 0$

To write $\frac{1}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{x}{x - 2} \cdot \frac{x - 4}{x - 4} + \frac{1}{x - 4} \cdot \frac{x - 2}{x - 2} - \frac{2}{(x - 4) (x - 2)} = 0$

Write each expression with a common denominator of $(x - 2) (x - 4)$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{x}{x - 2}$ and $\frac{x - 4}{x - 4}$ .

$\frac{x (x - 4)}{(x - 2) (x - 4)} + \frac{1}{x - 4} \cdot \frac{x - 2}{x - 2} - \frac{2}{(x - 4) (x - 2)} = 0$

Multiply $\frac{1}{x - 4}$ and $\frac{x - 2}{x - 2}$ .

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

Reorder the factors of $(x - 4) (x - 2)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Apply the distributive property.

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

Multiply $x$ by $x$ .

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

Move $- 4$ to the left of $x$ .

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

Add $- 4 x$ and $x$ .

$\frac{x^{2} - 3 x - 2}{(x - 2) (x - 4)} - \frac{2}{(x - 4) (x - 2)} = 0$

Combine into one fraction.

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Reorder the factors of $(x - 4) (x - 2)$ .

$\frac{x^{2} - 3 x - 2}{(x - 2) (x - 4)} - \frac{2}{(x - 2) (x - 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x^{2} - 3 x - 2 - 2}{(x - 2) (x - 4)} = 0$

Simplify the numerator.

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Subtract $2$ from $- 2$ .

$\frac{x^{2} - 3 x - 4}{(x - 2) (x - 4)} = 0$

Factor $x^{2} - 3 x - 4$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$\frac{(x - 4) (x + 1)}{(x - 2) (x - 4)} = 0$

Cancel the common factor of $x - 4$ .

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Cancel the common factor.

$\frac{(x - 4) (x + 1)}{(x - 2) (x - 4)} = 0$

Rewrite the expression.

$\frac{x + 1}{x - 2} = 0$

Factor $x^{2} - 6 x + 8$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$\frac{x}{x - 2} + \frac{1}{x - 4} - \frac{2}{(x - 4) (x - 2)} = 0$

To write $\frac{x}{x - 2}$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$\frac{x}{x - 2} \cdot \frac{x - 4}{x - 4} + \frac{1}{x - 4} - \frac{2}{(x - 4) (x - 2)} = 0$

To write $\frac{1}{x - 4}$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$\frac{x}{x - 2} \cdot \frac{x - 4}{x - 4} + \frac{1}{x - 4} \cdot \frac{x - 2}{x - 2} - \frac{2}{(x - 4) (x - 2)} = 0$

Write each expression with a common denominator of $(x - 2) (x - 4)$ , by multiplying each by an appropriate factor of $1$ .

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Multiply $\frac{x}{x - 2}$ and $\frac{x - 4}{x - 4}$ .

$\frac{x (x - 4)}{(x - 2) (x - 4)} + \frac{1}{x - 4} \cdot \frac{x - 2}{x - 2} - \frac{2}{(x - 4) (x - 2)} = 0$

Multiply $\frac{1}{x - 4}$ and $\frac{x - 2}{x - 2}$ .

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

Reorder the factors of $(x - 4) (x - 2)$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

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Apply the distributive property.

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

Multiply $x$ by $x$ .

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

Move $- 4$ to the left of $x$ .

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

Add $- 4 x$ and $x$ .

$\frac{x^{2} - 3 x - 2}{(x - 2) (x - 4)} - \frac{2}{(x - 4) (x - 2)} = 0$

Combine into one fraction.

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Reorder the factors of $(x - 4) (x - 2)$ .

$\frac{x^{2} - 3 x - 2}{(x - 2) (x - 4)} - \frac{2}{(x - 2) (x - 4)} = 0$

Combine the numerators over the common denominator.

$\frac{x^{2} - 3 x - 2 - 2}{(x - 2) (x - 4)} = 0$

Simplify the numerator.

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Subtract $2$ from $- 2$ .

$\frac{x^{2} - 3 x - 4}{(x - 2) (x - 4)} = 0$

Factor $x^{2} - 3 x - 4$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 4$ and whose sum is $- 3$ .

$- 4, 1$

Write the factored form using these integers.

$\frac{(x - 4) (x + 1)}{(x - 2) (x - 4)} = 0$

Cancel the common factor of $x - 4$ .

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Cancel the common factor.

$\frac{(x - 4) (x + 1)}{(x - 2) (x - 4)} = 0$

Rewrite the expression.

$\frac{x + 1}{x - 2} = 0$

Find the LCD of the terms in the equation.

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

$x - 2, 1$

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.

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

Not prime

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

$1$

The factor for $x - 2$ is $x - 2$ itself.

$(x - 2) = x - 2$

$(x - 2)$ occurs $1$ time.

The LCM of $x - 2$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x - 2$

Multiply each term by $x - 2$ and simplify.

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Multiply each term in $\frac{x + 1}{x - 2} = 0$ by $x - 2$ in order to remove all the denominators from the equation.

$\frac{x + 1}{x - 2} \cdot (x - 2) = 0 \cdot (x - 2)$

Cancel the common factor of $x - 2$ .

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Cancel the common factor.

$\frac{x + 1}{x - 2} \cdot (x - 2) = 0 \cdot (x - 2)$

Rewrite the expression.

$x + 1 = 0 \cdot (x - 2)$

Multiply $0$ by $x - 2$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Solve by Factoring x/(x-2)+1/(x-4)=2/(x^2-6x+8)

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