Solve for z 1/6(12z-18)=2z-3

$\frac{1}{6} (12 z - 18) = 2 z - 3$

Simplify $\frac{1}{6} \cdot (12 z - 18)$ .

Rewrite.

$0 + 0 + \frac{1}{6} \cdot (12 z - 18) = 2 z - 3$

Simplify by adding zeros.

$\frac{1}{6} \cdot (12 z - 18) = 2 z - 3$

Apply the distributive property.

$\frac{1}{6} (12 z) + \frac{1}{6} \cdot - 18 = 2 z - 3$

Cancel the common factor of $6$ .

Factor $6$ out of $12 z$ .

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

Cancel the common factor.

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

Rewrite the expression.

$2 z + \frac{1}{6} \cdot - 18 = 2 z - 3$

Cancel the common factor of $6$ .

Factor $6$ out of $- 18$ .

$2 z + \frac{1}{6} \cdot (6 (- 3)) = 2 z - 3$

Cancel the common factor.

$2 z + \frac{1}{6} \cdot (6 \cdot - 3) = 2 z - 3$

Rewrite the expression.

$2 z - 3 = 2 z - 3$

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

Subtract $2 z$ from both sides of the equation.

$2 z - 3 - 2 z = - 3$

Combine the opposite terms in $2 z - 3 - 2 z$ .

Subtract $2 z$ from $2 z$ .

$0 - 3 = - 3$

Subtract $3$ from $0$ .

$- 3 = - 3$

Since $- 3 = - 3$ , the equation will always be true for any value of $z$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$