To complete the table using the equation \( y = x - 1 \), we can substitute the values of \( x \) to find the corresponding \( y \) values.
Here’s the completed table:
\[ \begin{array}{|c|c|} \hline x & y \ \hline 2 & 1 \ 3 & 2 \ 4 & 3 \ 5 & 4 \ \hline \end{array} \]
So, the values of \( y \) are:
- When \( x = 2 \), \( y = 2 - 1 = 1 \)
- When \( x = 3 \), \( y = 3 - 1 = 2 \)
- When \( x = 4 \), \( y = 4 - 1 = 3 \)
- When \( x = 5 \), \( y = 5 - 1 = 4 \)