Solve for c 2c+3(c+3)=3(2c+3)

$2 c + 3 (c + 3) = 3 (2 c + 3)$

Simplify $2 c + 3 (c + 3)$ .

Simplify each term.

Apply the distributive property.

$2 c + 3 c + 3 \cdot 3 = 3 (2 c + 3)$

Multiply $3$ by $3$ .

$2 c + 3 c + 9 = 3 (2 c + 3)$

Add $2 c$ and $3 c$ .

$5 c + 9 = 3 (2 c + 3)$

Simplify $3 (2 c + 3)$ .

Apply the distributive property.

$5 c + 9 = 3 (2 c) + 3 \cdot 3$

Multiply.

Multiply $2$ by $3$ .

$5 c + 9 = 6 c + 3 \cdot 3$

Multiply $3$ by $3$ .

$5 c + 9 = 6 c + 9$

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

Subtract $6 c$ from both sides of the equation.

$5 c + 9 - 6 c = 9$

Subtract $6 c$ from $5 c$ .

$- c + 9 = 9$

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

Subtract $9$ from both sides of the equation.

$- c = 9 - 9$

Subtract $9$ from $9$ .

$- c = 0$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $c$ by $1$ .

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

Simplify the right side.

Divide $0$ by $- 1$ .

$c = 0$