Solve for c 9c-(q-2)=6c+2(c+1)

$9 c - (q - 2) = 6 c + 2 (c + 1)$

Step 1

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

$6 c + 2 (c + 1) = 9 c - (q - 2)$

Step 2

Simplify $6 c + 2 (c + 1)$ .

Simplify each term.

Apply the distributive property.

$6 c + 2 c + 2 \cdot 1 = 9 c - (q - 2)$

Multiply $2$ by $1$ .

$6 c + 2 c + 2 = 9 c - (q - 2)$

Add $6 c$ and $2 c$ .

$8 c + 2 = 9 c - (q - 2)$

Step 3

Simplify each term.

Apply the distributive property.

$8 c + 2 = 9 c - q - - 2$

Multiply $- 1$ by $- 2$ .

$8 c + 2 = 9 c - q + 2$

Step 4

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

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

$8 c + 2 - 9 c = - q + 2$

Subtract $9 c$ from $8 c$ .

$- c + 2 = - q + 2$

Step 5

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

Subtract $2$ from both sides of the equation.

$- c = - q + 2 - 2$

Combine the opposite terms in $- q + 2 - 2$ .

Subtract $2$ from $2$ .

$- c = - q + 0$

Add $- q$ and $0$ .

$- c = - q$

Step 6

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $c$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

$c = \frac{q}{1}$

Divide $q$ by $1$ .

$c = q$

Solve for c 9c-(q-2)=6c+2(c+1)

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