Certainly! Let's fill in the missing entries by solving the linear equation [tex]\(12x - 8y = 16\)[/tex] step-by-step.
### Fill in the Table:
1. First Missing Entry (when [tex]\(x\)[/tex] is missing and [tex]\(y = 0\)[/tex]):
Given:
[tex]\[
12x - 8(0) = 16 \implies 12x = 16 \implies x = \frac{16}{12} = \frac{4}{3}
\][/tex]
So, when [tex]\(y = 0\)[/tex], [tex]\(x = \frac{4}{3}\)[/tex].
2. Second Missing Entry (when [tex]\(x = 0\)[/tex]):
Given:
[tex]\[
12(0) - 8y = 16 \implies -8y = 16 \implies y = \frac{16}{-8} = -2
\][/tex]
So, when [tex]\(x = 0\)[/tex], [tex]\(y = -2\)[/tex].
3. Third Missing Entry (when [tex]\(x = 2\)[/tex]):
Given:
[tex]\[
12(2) - 8y = 16 \implies 24 - 8y = 16 \implies -8y = 16 - 24 \implies -8y = -8 \implies y = 1
\][/tex]
So, when [tex]\(x = 2\)[/tex], [tex]\(y = 1\)[/tex].
4. Fourth Missing Entry (when [tex]\(y\)[/tex] is missing and [tex]\(x = \frac{4}{3}\)[/tex]):
Checking the equation again:
[tex]\[
12 \left(\frac{4}{3}\right) - 8y = 16 \implies 16 - 8y = 16 \implies -8y = 0 \implies y = 0
\][/tex]
So, when [tex]\(x = \frac{4}{3}\)[/tex], [tex]\(y = 0\)[/tex].
### Final Table with Entries Filled:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
x & \frac{4}{3} & 0 & 2 & \frac{4}{3} \\ \hline
y & 0 & -2 & 1 & 0 \\ \hline
\end{tabular}
\][/tex]
Thus, the values of the missing entries in the table are:
[tex]\[
\begin{aligned}
&\text{When } x \text{ is missing and } y = 0, x = \frac{4}{3}. \\
&\text{When } x = 0, y = -2. \\
&\text{When } x = 2, y = 1. \\
&\text{When } y \text{ is missing and } x = \frac{4}{3}, y = 0.
\][/tex]
This is the completed table:
[tex]\[
\begin{tabular}{|l|l|l|l|l|l|}
\hline
x & \frac{4}{3} & 0 & 2 & \frac{4}{3} \\ \hline
y & 0 & -2 & 1 & 0 \\ \hline
\end{tabular}
\][/tex]
No calculations are necessary beyond these steps!
