Solve by Factoring 16/(8x-8)+1/8=2/(x-1)

$\frac{16}{8 x - 8} + \frac{1}{8} = \frac{2}{x - 1}$

Step 1

Subtract $\frac{2}{x - 1}$ from both sides of the equation.

$\frac{16}{8 x - 8} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Step 2

Simplify $\frac{16}{8 x - 8} + \frac{1}{8} - \frac{2}{x - 1}$ .

Cancel the common factor of $16$ and $8 x - 8$ .

Factor $8$ out of $16$ .

$\frac{8 (2)}{8 x - 8} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Cancel the common factors.

Factor $8$ out of $8 x$ .

$\frac{8 (2)}{8 (x) - 8} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Factor $8$ out of $- 8$ .

$\frac{8 \cdot 2}{8 (x) + 8 \cdot - 1} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Factor $8$ out of $8 (x) + 8 (- 1)$ .

$\frac{8 (2)}{8 (x - 1)} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Cancel the common factor.

$\frac{8 \cdot 2}{8 (x - 1)} + \frac{1}{8} - \frac{2}{x - 1} = 0$

Rewrite the expression.

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

Combine the opposite terms in $\frac{2}{x - 1} + \frac{1}{8} - \frac{2}{x - 1}$ .

Subtract $\frac{2}{x - 1}$ from $\frac{2}{x - 1}$ .

$0 + \frac{1}{8} = 0$

Add $0$ and $\frac{1}{8}$ .

$\frac{1}{8} = 0$

Step 3

Set the numerator equal to zero.

$1 = 0$

Step 4

Since $1 \neq 0$ , there are no solutions.

No solution