Solve for b 9+3b+2(b-1)=67

$9 + 3 b + 2 (b - 1) = 67$

Simplify $9 + 3 b + 2 (b - 1)$ .

Simplify each term.

Apply the distributive property.

$9 + 3 b + 2 b + 2 \cdot - 1 = 67$

Multiply $2$ by $- 1$ .

$9 + 3 b + 2 b - 2 = 67$

Simplify by adding terms.

Subtract $2$ from $9$ .

$3 b + 2 b + 7 = 67$

Add $3 b$ and $2 b$ .

$5 b + 7 = 67$

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

Subtract $7$ from both sides of the equation.

$5 b = 67 - 7$

Subtract $7$ from $67$ .

$5 b = 60$

Divide each term in $5 b = 60$ by $5$ and simplify.

Divide each term in $5 b = 60$ by $5$ .

$\frac{5 b}{5} = \frac{60}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 b}{5} = \frac{60}{5}$

Divide $b$ by $1$ .

$b = \frac{60}{5}$

Simplify the right side.

Divide $60$ by $5$ .

$b = 12$