Solve for z -(3z+4)=6z-3(3z+2)

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

Step 1

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

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

Step 2

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

Simplify each term.

Apply the distributive property.

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

Multiply $3$ by $- 3$ .

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

Multiply $- 3$ by $2$ .

$6 z - 9 z - 6 = - (3 z + 4)$

Subtract $9 z$ from $6 z$ .

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

Step 3

Simplify $- (3 z + 4)$ .

Apply the distributive property.

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

Multiply.

Multiply $3$ by $- 1$ .

$- 3 z - 6 = - 3 z - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$- 3 z - 6 = - 3 z - 4$

Step 4

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

Add $3 z$ to both sides of the equation.

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

Combine the opposite terms in $- 3 z - 6 + 3 z$ .

Add $- 3 z$ and $3 z$ .

$0 - 6 = - 4$

Subtract $6$ from $0$ .

$- 6 = - 4$

Step 5

Since $- 6 \neq - 4$ , there are no solutions.

No solution