\( \frac{x^{3}+3 y}{6}=\frac{z-y}{2} \) Given \( x=2 \) and \( y=3 \), solve for \( z \).

1. Substitute \( x = 2 \) and \( y = 3 \) into the equation: \[ \frac{2^3 + 3 \cdot 3}{6} = \frac{z - 3}{2} \] 2. Compute the values in the numerator on the left side: \[ 2^3 = 8 \quad \text{and} \quad 3 \cdot 3 = 9, \] so: \[ 8 + 9 = 17. \] The equation becomes: \[ \frac{17}{6} = \frac{z - 3}{2}. \] 3. Multiply both sides of the equation by 2 to eliminate the denominator on the right: \[ 2 \cdot \frac{17}{6} = z - 3. \] Simplify the left side: \[ \frac{34}{6} = \frac{17}{3} = z - 3. \] 4. Solve for \( z \) by adding 3 to both sides: \[ z = \frac{17}{3} + 3. \] 5. Express 3 as a fraction with denominator 3: \[ 3 = \frac{9}{3}. \] Thus: \[ z = \frac{17}{3} + \frac{9}{3} = \frac{26}{3}. \] The solution is: \[ z = \frac{26}{3}. \]