Solve by Factoring -(3z+4)=6z-3(3z+2)

$- (3 z + 4) = 6 z - 3 (3 z + 2)$

Step 1

Move all the expressions to the left side of the equation.

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

$- (3 z + 4) - 6 z = - 3 (3 z + 2)$

Add $3 (3 z + 2)$ to both sides of the equation.

$- (3 z + 4) - 6 z + 3 (3 z + 2) = 0$

Step 2

Simplify $- (3 z + 4) - 6 z + 3 (3 z + 2)$ .

Simplify each term.

Apply the distributive property.

$- (3 z) - 1 \cdot 4 - 6 z + 3 (3 z + 2) = 0$

Multiply $3$ by $- 1$ .

$- 3 z - 1 \cdot 4 - 6 z + 3 (3 z + 2) = 0$

Multiply $- 1$ by $4$ .

$- 3 z - 4 - 6 z + 3 (3 z + 2) = 0$

Apply the distributive property.

$- 3 z - 4 - 6 z + 3 (3 z) + 3 \cdot 2 = 0$

Multiply $3$ by $3$ .

$- 3 z - 4 - 6 z + 9 z + 3 \cdot 2 = 0$

Multiply $3$ by $2$ .

$- 3 z - 4 - 6 z + 9 z + 6 = 0$

Subtract $6 z$ from $- 3 z$ .

$- 9 z - 4 + 9 z + 6 = 0$

Combine the opposite terms in $- 9 z - 4 + 9 z + 6$ .

Add $- 9 z$ and $9 z$ .

$0 - 4 + 6 = 0$

Subtract $4$ from $0$ .

$- 4 + 6 = 0$

Add $- 4$ and $6$ .

$2 = 0$

Step 3

Since $2 \neq 0$ , there are no solutions.

No solution