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

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

Step 1

Move all the expressions to the left side of the equation.

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

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

Subtract $4$ from both sides of the equation.

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

Step 2

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

Find the common denominator.

Write $- 4$ as a fraction with denominator $1$ .

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

Multiply $\frac{- 4}{1}$ by $\frac{x - 3}{x - 3}$ .

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

Multiply $\frac{- 4}{1}$ by $\frac{x - 3}{x - 3}$ .

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

$\frac{x - 3 - 4 x - 4 \cdot - 3}{x - 3} = 0$

Multiply $- 4$ by $- 3$ .

$\frac{x - 3 - 4 x + 12}{x - 3} = 0$

Subtract $4 x$ from $x$ .

$\frac{- 3 x - 3 + 12}{x - 3} = 0$

Add $- 3$ and $12$ .

$\frac{- 3 x + 9}{x - 3} = 0$

Factor $3$ out of $- 3 x + 9$ .

Factor $3$ out of $- 3 x$ .

$\frac{3 (- x) + 9}{x - 3} = 0$

Factor $3$ out of $9$ .

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

Factor $3$ out of $3 (- x) + 3 (3)$ .

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

Cancel the common factor of $- x + 3$ and $x - 3$ .

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

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

Rewrite $3$ as $- 1 (- 3)$ .

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

Factor $- 1$ out of $- (x) - 1 (- 3)$ .

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

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

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

Cancel the common factor.

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

Divide $3 \cdot (- 1)$ by $1$ .

$3 \cdot (- 1) = 0$

Multiply $3$ by $- 1$ .

$- 3 = 0$

Step 3

Since $- 3 \neq 0$ , there are no solutions.

No solution

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

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