Solve for b 3(b+3)-4(-2+b)=19

$3 (b + 3) - 4 (- 2 + b) = 19$

Step 1

Simplify $3 (b + 3) - 4 (- 2 + b)$ .

Simplify each term.

Apply the distributive property.

$3 b + 3 \cdot 3 - 4 (- 2 + b) = 19$

Multiply $3$ by $3$ .

$3 b + 9 - 4 (- 2 + b) = 19$

Apply the distributive property.

$3 b + 9 - 4 \cdot - 2 - 4 b = 19$

Multiply $- 4$ by $- 2$ .

$3 b + 9 + 8 - 4 b = 19$

Simplify by adding terms.

Subtract $4 b$ from $3 b$ .

$- b + 9 + 8 = 19$

Add $9$ and $8$ .

$- b + 17 = 19$

Step 2

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

Subtract $17$ from both sides of the equation.

$- b = 19 - 17$

Subtract $17$ from $19$ .

$- b = 2$

Step 3

Divide each term in $- b = 2$ by $- 1$ and simplify.

Divide each term in $- b = 2$ by $- 1$ .

$\frac{- b}{- 1} = \frac{2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{b}{1} = \frac{2}{- 1}$

Divide $b$ by $1$ .

$b = \frac{2}{- 1}$

Simplify the right side.

Divide $2$ by $- 1$ .

$b = - 2$