Question

To find the solution to a system of linear equations, Verdita begins by creating equations for the two sets of data points below.

Data Set A
A 2-column table with 4 rows. Column 1 is labeled x with entries negative 1, 1, 5, 7. Column 2 is labeled y with entries negative 6, 2, 18, 26.
Data Set B
A 2-column table with 4 rows. Column 1 is labeled x with entries negative 5, negative 2, 0, 6. Column 2 is labeled y with entries negative 1, 2, 4, 10.

Which equations could Verdita use to represent the data sets?
Data Set A: y = 4 x minus 2
Data Set B: y = x + 4
Data Set A: y = 4 x minus 2
Data Set B: y = x minus 4
Data Set A: y = 4 x + 2
Data Set B: y = x minus 4
Data Set A: y = 4 x + 2
Data Set B: y = x + 4

Answer 1

To find the equations that represent the data from the two sets, we can analyze each data set by using the given points to derive the equations.

### Data Set A:
Points: (-1, -6), (1, 2), (5, 18), (7, 26)

1. To find the slope (m) of the line, we can use two points, e.g., (1, 2) and (5, 18):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{18 - 2}{5 - 1} = \frac{16}{4} = 4
\]

2. Now we can use the point-slope form \(y - y_1 = m(x - x_1)\) using the point (1, 2):
\[
y - 2 = 4(x - 1)
\]
Simplifying:
\[
y - 2 = 4x - 4 \Rightarrow y = 4x - 2
\]

So, the equation for Data Set A is:
\[
y = 4x - 2
\]

### Data Set B:
Points: (-5, -1), (-2, 2), (0, 4), (6, 10)

1. We can again use two points, e.g., (-5, -1) and (0, 4):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{0 - (-5)} = \frac{5}{5} = 1
\]

2. Now we can use the point-slope form with the point (0, 4):
\[
y - 4 = 1(x - 0)
\]
Simplifying:
\[
y = x + 4
\]

So, the equation for Data Set B is:
\[
y = x + 4
\]

### Conclusion:
- For Data Set A, the equation is \(y = 4x - 2\).
- For Data Set B, the equation is \(y = x + 4\).

Thus, the correct answer is:
**Data Set A: y = 4x - 2**
**Data Set B: y = x + 4**

Therefore, the correct option is:
**Data Set A: y = 4x - 2**
**Data Set B: y = x + 4**