Sure! To complete the table of values for the equation [tex]\( y = 5 - x \)[/tex], we need to substitute each [tex]\( x \)[/tex] value into the equation and find the corresponding [tex]\( y \)[/tex] value.
1. When [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 5 - 0 = 5
\][/tex]
2. When [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 5 - 1 = 4
\][/tex]
3. When [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 5 - 2 = 3
\][/tex]
With these calculations, the completed table of values is:
[tex]\[
\begin{tabular}{|l|l|l|l|}
\hline
$x$ & 0 & 1 & 2 \\
\hline
$y$ & 5 & 4 & 3 \\
\hline
\end{tabular}
\][/tex]
