Solve for y 8x-y=-8+6

$8 x - y = - 8 + 6$

Add $- 8$ and $6$ .

$8 x - y = - 2$

Subtract $8 x$ from both sides of the equation.

$- y = - 2 - 8 x$

Divide each term in $- y = - 2 - 8 x$ by $- 1$ and simplify.

Divide each term in $- y = - 2 - 8 x$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Divide $- 2$ by $- 1$ .

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

Move the negative one from the denominator of $\frac{- 8 x}{- 1}$ .

$y = 2 - 1 \cdot (- 8 x)$

Rewrite $- 1 \cdot (- 8 x)$ as $- (- 8 x)$ .

$y = 2 - (- 8 x)$

Multiply $- 8$ by $- 1$ .

$y = 2 + 8 x$