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

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

Step 1

Subtract $\frac{2}{x^{2} - 6 x + 8}$ from both sides of the equation.

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

Step 2

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

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

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 $1$ .

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

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

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

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

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.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $x$ by $x$ .

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

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

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

Add $- 4 x$ and $x$ .

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

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.

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 $x - 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

Set the numerator equal to zero.

$x + 1 = 0$

Step 4

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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