1. The problem is to adjust the number 3 by putting the denominator of the term \frac{3y-2}{y}. 2. The term given is \frac{3y-2}{y}, where the denominator is $y$. 3. To adjust the number 3 by putting the denominator $y$, we multiply 3 by $y$ in the denominator, which means writing 3 as \frac{3y}{y}. 4. This gives us \frac{3y}{y}, which has the same denominator as the term \frac{3y-2}{y}. 5. Therefore, the adjusted number 3 with the denominator $y$ is \frac{3y}{y}. Final answer: $$\frac{3y}{y}$$