Worksheet for writing linear equations from tables, featuring four problems with variable pairs.

We are given four tables, each showing a linear relationship between two variables. For each, we must write a linear equation that defines the dependent variable in terms of the independent variable.

A linear equation has the form:
> y = mx + b
where:
- `y` is the dependent variable,
- `x` is the independent variable,
- `m` is the slope (rate of change),
- `b` is the y-intercept (value of y when x = 0).

---

Problem 1: Define p in terms of k

Table:
```
k | p
0 | 16
3 | 15
6 | 14
9 | 13
```

- Independent variable: `k`
- Dependent variable: `p`

Step 1: Find the slope (m)
Use any two points. Let’s use (0, 16) and (3, 15):

> m = (change in p) / (change in k) = (15 - 16) / (3 - 0) = (-1)/3 = -1/3

Step 2: Find the y-intercept (b)
When k = 0, p = 16 → so b = 16

✔ Equation: p = (-1/3)k + 16

---

Problem 2: Define s in terms of p

Table:
```
p | s
0 | 3
1 | 5
2 | 7
3 | 9
```

- Independent variable: `p`
- Dependent variable: `s`

Step 1: Find slope (m)
Use (0, 3) and (1, 5):

> m = (5 - 3)/(1 - 0) = 2/1 = 2

Step 2: Find y-intercept (b)
When p = 0, s = 3 → b = 3

✔ Equation: s = 2p + 3

---

Problem 3: Define d in terms of a

Table:
```
a | d
0 | 0
4 | 5
8 | 10
12| 15
```

- Independent variable: `a`
- Dependent variable: `d`

Step 1: Find slope (m)
Use (0, 0) and (4, 5):

> m = (5 - 0)/(4 - 0) = 5/4

Step 2: Find y-intercept (b)
When a = 0, d = 0 → b = 0

✔ Equation: d = (5/4)a

*(You can also write it as d = 1.25a)*

---

Problem 4: Define k in terms of c

Table:
```
c | k
0 | 17
4 | 14
8 | 11
12| 8
```

- Independent variable: `c`
- Dependent variable: `k`

Step 1: Find slope (m)
Use (0, 17) and (4, 14):

> m = (14 - 17)/(4 - 0) = (-3)/4 = -3/4

Step 2: Find y-intercept (b)
When c = 0, k = 17 → b = 17

✔ Equation: k = (-3/4)c + 17

---

## ✔ Final Answers:

1. p = -⅓k + 16
2. s = 2p + 3
3. d = ⁵⁄₄a
4. k = -¾c + 17

These equations correctly model the linear relationships shown in each table.