Solve for z 6(z+2)=3(z+1)-18

$6 (z + 2) = 3 (z + 1) - 18$

Simplify $6 (z + 2)$ .

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Apply the distributive property.

$6 z + 6 \cdot 2 = 3 (z + 1) - 18$

Multiply $6$ by $2$ .

$6 z + 12 = 3 (z + 1) - 18$

Simplify $3 (z + 1) - 18$ .

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Simplify each term.

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Apply the distributive property.

$6 z + 12 = 3 z + 3 \cdot 1 - 18$

Multiply $3$ by $1$ .

$6 z + 12 = 3 z + 3 - 18$

Subtract $18$ from $3$ .

$6 z + 12 = 3 z - 15$

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

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Subtract $3 z$ from both sides of the equation.

$6 z + 12 - 3 z = - 15$

Subtract $3 z$ from $6 z$ .

$3 z + 12 = - 15$

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

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Subtract $12$ from both sides of the equation.

$3 z = - 15 - 12$

Subtract $12$ from $- 15$ .

$3 z = - 27$

Divide each term by $3$ and simplify.

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Divide each term in $3 z = - 27$ by $3$ .

$\frac{3 z}{3} = \frac{- 27}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 z}{3} = \frac{- 27}{3}$

Divide $z$ by $1$ .

$z = \frac{- 27}{3}$

Divide $- 27$ by $3$ .

$z = - 9$

Solve for z 6(z+2)=3(z+1)-18

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