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

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UpStudy AI · Accepted Answer

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:
$$
z = \frac{26}{3}.
$$
$$ z = \frac{26}{3} $$

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

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