1. The problem provides two tables of values with variables $x$ and $y$ and asks to analyze or find a relationship between $x$ and $y$ for each table. 2. To find the relationship, we can check if $y$ depends linearly on $x$ by calculating the slope $m$ using the formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ and then find the equation of the line $y = mx + b$. 3. For the first table: Points: $(-2,3)$ and $(1,2)$ Calculate slope: $$m = \frac{2 - 3}{1 - (-2)} = \frac{-1}{3} = -\frac{1}{3}$$ Find intercept $b$ using point $(-2,3)$: $$3 = -\frac{1}{3} \times (-2) + b \Rightarrow 3 = \frac{2}{3} + b \Rightarrow b = 3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3}$$ Equation: $$y = -\frac{1}{3}x + \frac{7}{3}$$ 4. Verify with other points: At $x=4$: $$y = -\frac{1}{3} \times 4 + \frac{7}{3} = -\frac{4}{3} + \frac{7}{3} = 1$$ Matches given $y=1$. At $x=8$: $$y = -\frac{1}{3} \times 8 + \frac{7}{3} = -\frac{8}{3} + \frac{7}{3} = -\frac{1}{3}$$ Matches given $y=-\frac{1}{3}$. At $x=11$: $$y = -\frac{1}{3} \times 11 + \frac{7}{3} = -\frac{11}{3} + \frac{7}{3} = -\frac{4}{3}$$ Matches given $y=-\frac{4}{3}$. 5. For the second table: Points: $(-3,-2)$ and $(5,2)$ Calculate slope: $$m = \frac{2 - (-2)}{5 - (-3)} = \frac{4}{8} = \frac{1}{2}$$ Find intercept $b$ using point $(-3,-2)$: $$-2 = \frac{1}{2} \times (-3) + b \Rightarrow -2 = -\frac{3}{2} + b \Rightarrow b = -2 + \frac{3}{2} = -\frac{4}{2} + \frac{3}{2} = -\frac{1}{2}$$ Equation: $$y = \frac{1}{2}x - \frac{1}{2}$$ 6. Verify with other points: At $x=7$: $$y = \frac{1}{2} \times 7 - \frac{1}{2} = \frac{7}{2} - \frac{1}{2} = 3$$ Matches given $y=3$. At $x=10$: $$y = \frac{1}{2} \times 10 - \frac{1}{2} = 5 - \frac{1}{2} = 4.5$$ Matches given $y=4.5$. At $x=14$: $$y = \frac{1}{2} \times 14 - \frac{1}{2} = 7 - \frac{1}{2} = 6.5$$ Matches given $y=6.5$. Final answers: - First table equation: $$y = -\frac{1}{3}x + \frac{7}{3}$$ - Second table equation: $$y = \frac{1}{2}x - \frac{1}{2}$$