Solve for c 7c+5=9(c-3)

$7 c + 5 = 9 (c - 3)$

Simplify $9 (c - 3)$ .

Rewrite.

$7 c + 5 = 0 + 0 + 9 (c - 3)$

Simplify by adding zeros.

$7 c + 5 = 9 (c - 3)$

Apply the distributive property.

$7 c + 5 = 9 c + 9 \cdot - 3$

Multiply $9$ by $- 3$ .

$7 c + 5 = 9 c - 27$

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

Subtract $9 c$ from both sides of the equation.

$7 c + 5 - 9 c = - 27$

Subtract $9 c$ from $7 c$ .

$- 2 c + 5 = - 27$

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

Subtract $5$ from both sides of the equation.

$- 2 c = - 27 - 5$

Subtract $5$ from $- 27$ .

$- 2 c = - 32$

Divide each term in $- 2 c = - 32$ by $- 2$ and simplify.

Divide each term in $- 2 c = - 32$ by $- 2$ .

$\frac{- 2 c}{- 2} = \frac{- 32}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 c}{- 2} = \frac{- 32}{- 2}$

Divide $c$ by $1$ .

$c = \frac{- 32}{- 2}$

Simplify the right side.

Divide $- 32$ by $- 2$ .

$c = 16$