Sure, let's fill in the values for the table using the equation [tex]\( x + 2y = 1 \)[/tex].
First, we need to express [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex]. We can rearrange the given equation:
[tex]\[ x + 2y = 1 \][/tex]
[tex]\[ 2y = 1 - x \][/tex]
[tex]\[ y = \frac{1 - x}{2} \][/tex]
Now, we will use this function [tex]\( y = \frac{1 - x}{2} \)[/tex] to find the corresponding [tex]\( y \)[/tex] values (denoted as [tex]\( f(x) \)[/tex]) for the given [tex]\( x \)[/tex] values.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = \frac{1 - (-2)}{2} = \frac{1 + 2}{2} = \frac{3}{2} = 1.5 \][/tex]
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{1 - 0}{2} = \frac{1}{2} = 0.5 \][/tex]
3. For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{1 - 3}{2} = \frac{-2}{2} = -1 \][/tex]
So, the completed table is:
\begin{tabular}{|l|l|l|}
\hline
[tex]$x$[/tex] & & [tex]$f(x)$[/tex] \\
\hline
-2 & & 1.5 \\
\hline
0 & & 0.5 \\
\hline
3 & & -1.0 \\
\hline
\end{tabular}
