Solve for w y-aw=2w-1

$y - a w = 2 w - 1$

Subtract $2 w$ from both sides of the equation.

$y - a w - 2 w = - 1$

Subtract $y$ from both sides of the equation.

$- a w - 2 w = - 1 - y$

Factor $w$ out of $- a w - 2 w$ .

Click for detailed explanation...

Factor $w$ out of $- a w$ .

$w (- a) - 2 w = - 1 - y$

Factor $w$ out of $- 2 w$ .

$w (- a) + w \cdot - 2 = - 1 - y$

Factor $w$ out of $w (- a) + w \cdot - 2$ .

$w (- a - 2) = - 1 - y$

Divide each term in $w (- a - 2) = - 1 - y$ by $- a - 2$ and simplify.

Click for detailed explanation...

Divide each term in $w (- a - 2) = - 1 - y$ by $- a - 2$ .

$\frac{w (- a - 2)}{- a - 2} = \frac{- 1}{- a - 2} + \frac{- y}{- a - 2}$

Simplify the left side.

Click for detailed explanation...

Cancel the common factor of $- a - 2$ .

Click for detailed explanation...

Cancel the common factor.

$\frac{w (- a - 2)}{- a - 2} = \frac{- 1}{- a - 2} + \frac{- y}{- a - 2}$

Divide $w$ by $1$ .

$w = \frac{- 1}{- a - 2} + \frac{- y}{- a - 2}$

Simplify the right side.

Click for detailed explanation...

Simplify each term.

Click for detailed explanation...

Move the negative in front of the fraction.

$w = - \frac{1}{- a - 2} + \frac{- y}{- a - 2}$

Move the negative in front of the fraction.

$w = - \frac{1}{- a - 2} - \frac{y}{- a - 2}$

Simplify terms.

Click for detailed explanation...

Combine the numerators over the common denominator.

$w = \frac{- 1 - y}{- a - 2}$

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

$w = \frac{- 1 (1) - y}{- a - 2}$

Factor $- 1$ out of $- y$ .

$w = \frac{- 1 (1) - (y)}{- a - 2}$

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

$w = \frac{- 1 (1 + y)}{- a - 2}$

Factor $- 1$ out of $- a$ .

$w = \frac{- 1 (1 + y)}{- (a) - 2}$

Rewrite $- 2$ as $- 1 (2)$ .

$w = \frac{- 1 (1 + y)}{- (a) - 1 (2)}$

Factor $- 1$ out of $- (a) - 1 (2)$ .

$w = \frac{- 1 (1 + y)}{- (a + 2)}$

Rewrite $- (a + 2)$ as $- 1 (a + 2)$ .

$w = \frac{- 1 (1 + y)}{- 1 (a + 2)}$

Cancel the common factor.

$w = \frac{- 1 (1 + y)}{- 1 (a + 2)}$

Rewrite the expression.

$w = \frac{1 + y}{a + 2}$