Sure, let's find the solutions to the equation [tex]\( y = 0.5x - 2 \)[/tex] for the given values of [tex]\( x \)[/tex].
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[
y = 0.5(-2) - 2 = -1 - 2 = -3.0
\][/tex]
Therefore, when [tex]\( x = -2 \)[/tex], [tex]\( y = -3.0 \)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[
y = 0.5(-1) - 2 = -0.5 - 2 = -2.5
\][/tex]
Therefore, when [tex]\( x = -1 \)[/tex], [tex]\( y = -2.5 \)[/tex].
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[
y = 0.5(0) - 2 = 0 - 2 = -2.0
\][/tex]
Therefore, when [tex]\( x = 0 \)[/tex], [tex]\( y = -2.0 \)[/tex].
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[
y = 0.5(1) - 2 = 0.5 - 2 = -1.5
\][/tex]
Therefore, when [tex]\( x = 1 \)[/tex], [tex]\( y = -1.5 \)[/tex].
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = 0.5(2) - 2 = 1 - 2 = -1.0
\][/tex]
Therefore, when [tex]\( x = 2 \)[/tex], [tex]\( y = -1.0 \)[/tex].
Now, we can fill in the table with the values we calculated:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
-2 & -3.0 \\
\hline
-1 & -2.5 \\
\hline
0 & -2.0 \\
\hline
1 & -1.5 \\
\hline
2 & -1.0 \\
\hline
\end{tabular}
\][/tex]
So, the table is complete with the corresponding [tex]\( y \)[/tex] values for the given [tex]\( x \)[/tex] values.
