Solve for c 4(3c-2)=12(c+2)

$4 (3 c - 2) = 12 (c + 2)$

Step 1

Simplify $4 (3 c - 2)$ .

Rewrite.

$0 + 0 + 4 (3 c - 2) = 12 (c + 2)$

Simplify by adding zeros.

$4 (3 c - 2) = 12 (c + 2)$

Apply the distributive property.

$4 (3 c) + 4 \cdot - 2 = 12 (c + 2)$

Multiply.

Multiply $3$ by $4$ .

$12 c + 4 \cdot - 2 = 12 (c + 2)$

Multiply $4$ by $- 2$ .

$12 c - 8 = 12 (c + 2)$

Step 2

Simplify $12 (c + 2)$ .

Apply the distributive property.

$12 c - 8 = 12 c + 12 \cdot 2$

Multiply $12$ by $2$ .

$12 c - 8 = 12 c + 24$

Step 3

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

Subtract $12 c$ from both sides of the equation.

$12 c - 8 - 12 c = 24$

Combine the opposite terms in $12 c - 8 - 12 c$ .

Subtract $12 c$ from $12 c$ .

$0 - 8 = 24$

Subtract $8$ from $0$ .

$- 8 = 24$

Step 4

Since $- 8 \neq 24$ , there are no solutions.

No solution