To find solutions to the linear equation [tex]\(3x - 4y = 12\)[/tex], we need to determine the corresponding [tex]\(y\)[/tex]-values for given [tex]\(x\)[/tex]-values. Here is a step-by-step calculation for each [tex]\(x\)[/tex]-value provided in the table:
1. For [tex]\(x = 0\)[/tex]:
[tex]\[
3(0) - 4y = 12
\][/tex]
[tex]\[
0 - 4y = 12
\][/tex]
[tex]\[
-4y = 12
\][/tex]
[tex]\[
y = \frac{12}{-4}
\][/tex]
[tex]\[
y = -3
\][/tex]
So, when [tex]\(x = 0\)[/tex], [tex]\(y = -3\)[/tex].
2. For [tex]\(x = 8\)[/tex]:
[tex]\[
3(8) - 4y = 12
\][/tex]
[tex]\[
24 - 4y = 12
\][/tex]
[tex]\[
-4y = 12 - 24
\][/tex]
[tex]\[
-4y = -12
\][/tex]
[tex]\[
y = \frac{-12}{-4}
\][/tex]
[tex]\[
y = 3
\][/tex]
So, when [tex]\(x = 8\)[/tex], [tex]\(y = 3\)[/tex].
3. For [tex]\(x = 12\)[/tex]:
[tex]\[
3(12) - 4y = 12
\][/tex]
[tex]\[
36 - 4y = 12
\][/tex]
[tex]\[
-4y = 12 - 36
\][/tex]
[tex]\[
-4y = -24
\][/tex]
[tex]\[
y = \frac{-24}{-4}
\][/tex]
[tex]\[
y = 6
\][/tex]
So, when [tex]\(x = 12\)[/tex], [tex]\(y = 6\)[/tex].
Now we can complete the table with the calculated [tex]\(y\)[/tex]-values:
[tex]\[
\begin{tabular}{|c|c|}
\hline
$x$ & $y$ \\
\hline
0 & -3 \\
\hline
8 & 3 \\
\hline
12 & 6 \\
\hline
\end{tabular}
\][/tex]
