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

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

Step 1

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

Subtract $4$ from both sides of the equation.

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

Add $\frac{8}{x + 1}$ to both sides of the equation.

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

Step 2

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

Find the common denominator.

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

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

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

$\frac{8 x}{x + 1} + \frac{- 4}{1} \cdot \frac{x + 1}{x + 1} + \frac{8}{x + 1} = 0$

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

$\frac{8 x}{x + 1} + \frac{- 4 (x + 1)}{x + 1} + \frac{8}{x + 1} = 0$

Combine the numerators over the common denominator.

$\frac{8 x - 4 (x + 1) + 8}{x + 1} = 0$

Simplify each term.

Apply the distributive property.

$\frac{8 x - 4 x - 4 \cdot 1 + 8}{x + 1} = 0$

Multiply $- 4$ by $1$ .

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

Subtract $4 x$ from $8 x$ .

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

Add $- 4$ and $8$ .

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

Factor $4$ out of $4 x + 4$ .

Factor $4$ out of $4 x$ .

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

Factor $4$ out of $4$ .

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

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

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Divide $4$ by $1$ .

$4 = 0$

Step 3

Since $4 \neq 0$ , there are no solutions.

No solution