Solve for c 2c-3=2(6-c)+7c

$2 c - 3 = 2 (6 - c) + 7 c$

Since $c$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 (6 - c) + 7 c = 2 c - 3$

Simplify $2 (6 - c) + 7 c$ .

Simplify each term.

Apply the distributive property.

$2 \cdot 6 + 2 (- c) + 7 c = 2 c - 3$

Multiply $2$ by $6$ .

$12 + 2 (- c) + 7 c = 2 c - 3$

Multiply $- 1$ by $2$ .

$12 - 2 c + 7 c = 2 c - 3$

Add $- 2 c$ and $7 c$ .

$12 + 5 c = 2 c - 3$

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

Subtract $2 c$ from both sides of the equation.

$12 + 5 c - 2 c = - 3$

Subtract $2 c$ from $5 c$ .

$12 + 3 c = - 3$

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

Subtract $12$ from both sides of the equation.

$3 c = - 3 - 12$

Subtract $12$ from $- 3$ .

$3 c = - 15$

Divide each term in $3 c = - 15$ by $3$ and simplify.

Divide each term in $3 c = - 15$ by $3$ .

$\frac{3 c}{3} = \frac{- 15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 c}{3} = \frac{- 15}{3}$

Divide $c$ by $1$ .

$c = \frac{- 15}{3}$

Simplify the right side.

Divide $- 15$ by $3$ .

$c = - 5$