1. The problem asks to write an equation that models the relationship between the two columns of numbers in each table. 2. We will use the linear equation form $$y = mx + b$$ where $m$ is the slope (rate of change) and $b$ is the y-intercept (value when the independent variable is 0). 3. For each table, calculate the slope $m$ using two points: $$m = \frac{\text{change in dependent variable}}{\text{change in independent variable}} = \frac{y_2 - y_1}{x_2 - x_1}$$. 4. Then find $b$ by substituting one point into the equation and solving for $b$. **a)** Table with $x$ and $y$: - Points: $(0,13)$ and $(1,16)$ - Slope: $$m = \frac{16 - 13}{1 - 0} = \frac{3}{1} = 3$$ - Find $b$ using point $(0,13)$: $$13 = 3 \times 0 + b \Rightarrow b = 13$$ - Equation: $$y = 3x + 13$$ **b)** Table with $r$ and $p$: - Points: $(0,17)$ and $(1,24)$ - Slope: $$m = \frac{24 - 17}{1 - 0} = \frac{7}{1} = 7$$ - Find $b$ using point $(0,17)$: $$17 = 7 \times 0 + b \Rightarrow b = 17$$ - Equation: $$p = 7r + 17$$ **c)** Table with $k$ and $t$: - Points: $(1,-2)$ and $(2,1)$ - Slope: $$m = \frac{1 - (-2)}{2 - 1} = \frac{3}{1} = 3$$ - Find $b$ using point $(1,-2)$: $$-2 = 3 \times 1 + b \Rightarrow b = -2 - 3 = -5$$ - Equation: $$t = 3k - 5$$ **d)** Table with $f$ and $w$: - Points: $(1,-1)$ and $(2,-3)$ - Slope: $$m = \frac{-3 - (-1)}{2 - 1} = \frac{-2}{1} = -2$$ - Find $b$ using point $(1,-1)$: $$-1 = -2 \times 1 + b \Rightarrow b = -1 + 2 = 1$$ - Equation: $$w = -2f + 1$$