Question

Julius records the following set of tables during an experiment. Which system of equations could be used to represent the two tables?


Table 1
x
1
2
3
4
5
y
3
4
5
6
7

Table 2
x
1
2
3
4
5
y
6
3
0
-3-6



a.y = x + 2
y = −3x + 9

b. y = 2x + 1
y = 9x − 3

c.y = x + 2
y = 3x + 3

d.y = 2x + 1
y = 3x + 3

Answer 1

To find the correct system of equations to represent the two tables, we need to determine the equations for the lines represented by the data in each table.

**Table 1:**
```
x | y
-------
1 | 3
2 | 4
3 | 5
4 | 6
5 | 7
```

By examining Table 1, we can see that as \( x \) increases by 1, \( y \) increases by 1 as well. This indicates a linear relationship with a slope of 1. The y-intercept when \( x = 0 \) appears to be 2 (since when \( x = 1 \), \( y = 3 \)). Thus, the equation for Table 1 can be derived as:

\[
y = x + 2
\]

**Table 2:**
```
x | y
-------
1 | 6
2 | 3
3 | 0
4 | -3
5 | -6
```

In Table 2, as \( x \) increases by 1, \( y \) decreases by 3. Therefore, the slope is -3. To find the y-intercept, we can use the point \( (1, 6) \) and plug it into the slope-intercept form \( y = mx + b \):

\[
6 = -3(1) + b \implies b = 6 + 3 = 9
\]

Thus, the equation for Table 2 is:

\[
y = -3x + 9
\]

Now we have the equations:
1. From Table 1: \( y = x + 2 \)
2. From Table 2: \( y = -3x + 9 \)

Looking at the answer choices, we find that:

- **Answer choice (a) is correct:**
\[
y = x + 2 \quad (Table 1)
\]
\[
y = -3x + 9 \quad (Table 2)
\]

Thus, the answer is **(a)**: \( y = x + 2 \) and \( y = -3x + 9 \).