Solve by Factoring 1/x+1/(x-4)=(x-3)/(x-4)

$\frac{1}{x} + \frac{1}{x - 4} = \frac{x - 3}{x - 4}$

Step 1

Subtract $\frac{x - 3}{x - 4}$ from both sides of the equation.

$\frac{1}{x} + \frac{1}{x - 4} - \frac{x - 3}{x - 4} = 0$

Step 2

Simplify $\frac{1}{x} + \frac{1}{x - 4} - \frac{x - 3}{x - 4}$ .

Combine the numerators over the common denominator.

$\frac{1}{x} + \frac{1 - (x - 3)}{x - 4} = 0$

Simplify each term.

Apply the distributive property.

$\frac{1}{x} + \frac{1 - x - - 3}{x - 4} = 0$

Multiply $- 1$ by $- 3$ .

$\frac{1}{x} + \frac{1 - x + 3}{x - 4} = 0$

Add $1$ and $3$ .

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

Cancel the common factor of $- x + 4$ and $x - 4$ .

Factor $- 1$ out of $- x$ .

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

Rewrite $4$ as $- 1 (- 4)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 4)$ .

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

Rewrite $- (x - 4)$ as $- 1 (x - 4)$ .

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

Cancel the common factor.

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

Divide $- 1$ by $1$ .

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

Step 3

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

$\frac{1}{x} - 1 \cdot \frac{x}{x} = 0$

Step 4

Combine $- 1$ and $\frac{x}{x}$ .

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Set the numerator equal to zero.

$1 - x = 0$

Step 7

Solve the equation for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

Solve by Factoring 1/x+1/(x-4)=(x-3)/(x-4)

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