Certainly! Let's solve the given equation [tex]\(4y = -x + 12\)[/tex] step-by-step for the given [tex]\(x\)[/tex] values.
1. Start with the equation: [tex]\(4y = -x + 12\)[/tex]
2. Isolate [tex]\(y\)[/tex] by dividing both sides by 4:
[tex]\[
y = \frac{-x + 12}{4}
\][/tex]
Now, we will evaluate [tex]\(y\)[/tex] for each given [tex]\(x\)[/tex] value:
- For [tex]\(x = -8\)[/tex]:
[tex]\[
y = \frac{-(-8) + 12}{4} = \frac{8 + 12}{4} = \frac{20}{4} = 5.0
\][/tex]
- For [tex]\(x = -4\)[/tex]:
[tex]\[
y = \frac{-(-4) + 12}{4} = \frac{4 + 12}{4} = \frac{16}{4} = 4.0
\][/tex]
- For [tex]\(x = 0\)[/tex]:
[tex]\[
y = \frac{-(0) + 12}{4} = \frac{12}{4} = 3.0
\][/tex]
- For [tex]\(x = 4\)[/tex]:
[tex]\[
y = \frac{-(4) + 12}{4} = \frac{-4 + 12}{4} = \frac{8}{4} = 2.0
\][/tex]
- For [tex]\(x = 8\)[/tex]:
[tex]\[
y = \frac{-(8) + 12}{4} = \frac{-8 + 12}{4} = \frac{4}{4} = 1.0
\][/tex]
Let's fill in the table with these results:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & -8 & -4 & 0 & 4 & 8 \\
\hline
$y$ & 5.0 & 4.0 & 3.0 & 2.0 & 1.0 \\
\hline
\end{tabular}
\][/tex]
Therefore, the values of [tex]\(y\)[/tex] corresponding to the given [tex]\(x\)[/tex] values are [tex]\(5.0, 4.0, 3.0, 2.0,\)[/tex] and [tex]\(1.0\)[/tex], respectively.
