Solve for b 5(-b-2)=-4b-9-1

$5 (- b - 2) = - 4 b - 9 - 1$

Simplify $5 (- b - 2)$ .

Rewrite.

$0 + 0 + 5 (- b - 2) = - 4 b - 9 - 1$

Simplify by adding zeros.

$5 (- b - 2) = - 4 b - 9 - 1$

Apply the distributive property.

$5 (- b) + 5 \cdot - 2 = - 4 b - 9 - 1$

Multiply.

Multiply $- 1$ by $5$ .

$- 5 b + 5 \cdot - 2 = - 4 b - 9 - 1$

Multiply $5$ by $- 2$ .

$- 5 b - 10 = - 4 b - 9 - 1$

Subtract $1$ from $- 9$ .

$- 5 b - 10 = - 4 b - 10$

Move all terms containing $b$ to the left side of the equation.

Add $4 b$ to both sides of the equation.

$- 5 b - 10 + 4 b = - 10$

Add $- 5 b$ and $4 b$ .

$- b - 10 = - 10$

Move all terms not containing $b$ to the right side of the equation.

Add $10$ to both sides of the equation.

$- b = - 10 + 10$

Add $- 10$ and $10$ .

$- b = 0$

Divide each term in $- b = 0$ by $- 1$ and simplify.

Divide each term in $- b = 0$ by $- 1$ .

$\frac{- b}{- 1} = \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{b}{1} = \frac{0}{- 1}$

Divide $b$ by $1$ .

$b = \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$b = 0$