Solve for a -ab+a=b-1

$- a b + a = b - 1$

Step 1

Factor $a$ out of $- a b + a$ .

Factor $a$ out of $- a b$ .

$a (- 1 b) + a = b - 1$

Raise $a$ to the power of $1$ .

$a (- 1 b) + a = b - 1$

Factor $a$ out of $a^{1}$ .

$a (- 1 b) + a \cdot 1 = b - 1$

Factor $a$ out of $a (- 1 b) + a \cdot 1$ .

$a (- 1 b + 1) = b - 1$

Step 2

Rewrite $- 1 b$ as $- b$ .

$a (- b + 1) = b - 1$

Step 3

Divide each term in $a (- b + 1) = b - 1$ by $- b + 1$ and simplify.

Divide each term in $a (- b + 1) = b - 1$ by $- b + 1$ .

$\frac{a (- b + 1)}{- b + 1} = \frac{b}{- b + 1} + \frac{- 1}{- b + 1}$

Simplify the left side.

Cancel the common factor of $- b + 1$ .

Cancel the common factor.

$\frac{a (- b + 1)}{- b + 1} = \frac{b}{- b + 1} + \frac{- 1}{- b + 1}$

Divide $a$ by $1$ .

$a = \frac{b}{- b + 1} + \frac{- 1}{- b + 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$a = \frac{b - 1}{- b + 1}$

Cancel the common factor of $b - 1$ and $- b + 1$ .

Factor $- 1$ out of $b$ .

$a = \frac{- 1 (- b) - 1}{- b + 1}$

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

$a = \frac{- 1 (- b) - 1 (1)}{- b + 1}$

Factor $- 1$ out of $- 1 (- b) - 1 (1)$ .

$a = \frac{- 1 (- b + 1)}{- b + 1}$

Cancel the common factor.

$a = \frac{- 1 (- b + 1)}{- b + 1}$

Divide $- 1$ by $1$ .

$a = - 1$