To solve this problem, let's first determine the equations that fit the tables of values provided.
### For the first table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 2 \\
\hline
-4 & 0 \\
\hline
6 & 5 \\
\hline
-6 & -1 \\
\hline
\end{array}
\][/tex]
Using these points, we can find the best-fit line [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the intercept.
Slope [tex]\(m\)[/tex] is calculated using the formula:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Using two points (0, 2) and (-4, 0):
[tex]\[
m = \frac{0 - 2}{-4 - 0} = \frac{-2}{-4} = \frac{1}{2}
\][/tex]
Intercept [tex]\(b\)[/tex]:
Using the slope and one of the points, say (0, 2):
[tex]\[
y = mx + b \implies 2 = \frac{1}{2}(0) + b \implies b = 2
\][/tex]
Thus, the equation is:
[tex]\[
y = \frac{1}{2}x + 2
\][/tex]
### For the second table:
[tex]\[
\begin{array}{|c|c|}
\hline
x & y \\
\hline
0 & 1 \\
\hline
-2 & 0 \\
\hline
-4 & -1 \\
\hline
2 & 2 \\
\hline
\end{array}
\][/tex]
Using these points, we can find the best-fit line [tex]\(y = mx + b\)[/tex].
Slope [tex]\(m\)[/tex] is calculated:
Using points (0, 1) and (-2, 0):
[tex]\[
m = \frac{0 - 1}{-2 - 0} = \frac{-1}{-2} = \frac{1}{2}
\][/tex]
Intercept [tex]\(b\)[/tex]:
Using the slope and one of the points, say (0, 1):
[tex]\[
y = mx + b \implies 1 = \frac{1}{2}(0) + b \implies b = 1
\][/tex]
Thus, the equation is:
[tex]\[
y = \frac{1}{2}x + 1
\][/tex]
### Summary of linear equations:
1. [tex]\(y = \frac{1}{2}x + 2\)[/tex]
2. [tex]\(y = \frac{1}{2}x + 1\)[/tex]
Comparing these equations with the options:
A. [tex]\(y = -\frac{1}{2}x - 2\)[/tex] and [tex]\(y = -\frac{1}{2}x + 1\)[/tex]
B. [tex]\(y = -x - 2\)[/tex] and [tex]\(y = -x + 1\)[/tex]
C. [tex]\(y = \frac{1}{2}x + 2\)[/tex] and [tex]\(y = \frac{1}{2}x + 1\)[/tex]
D. [tex]\(y = x + 2\)[/tex] and [tex]\(y = x + 1\)[/tex]
Clearly, the equations match Option C:
[tex]\[
y = \frac{1}{2}x + 2 \quad \text{and} \quad y = \frac{1}{2}x + 1
\][/tex]
Therefore, the correct answer is C.
