1. **State the problem:** We need to fill in the table of values for each equation and graph the lines. 2. **Equation 1:** $-4x + 3y = -1$ - Solve for $y$: $$3y = 4x - 1$$ $$y = \frac{4x - 1}{3}$$ - Choose values for $x$ and find corresponding $y$: - For $x=0$: $$y = \frac{4(0) - 1}{3} = \frac{-1}{3}$$ - For $x=1$: $$y = \frac{4(1) - 1}{3} = \frac{3}{3} = 1$$ - For $x=-1$: $$y = \frac{4(-1) - 1}{3} = \frac{-4 - 1}{3} = \frac{-5}{3}$$ 3. **Equation 2:** $\frac{1}{3}y = -\frac{2}{3}$ - Multiply both sides by 3: $$\cancel{3} \times \frac{1}{\cancel{3}} y = -\frac{2}{3} \times 3$$ $$y = -2$$ - This is a horizontal line where $y$ is always $-2$ regardless of $x$. 4. **Equation 3:** $3x - 12 = 0$ - Solve for $x$: $$3x = 12$$ $$x = \frac{12}{3} = 4$$ - This is a vertical line where $x$ is always $4$ regardless of $y$. 5. **Summary of tables:** | Equation | $x$ | $y$ | |----------|-----|-----| | 1 | -1 | $-\frac{5}{3}$ | | | 0 | $-\frac{1}{3}$ | | | 1 | 1 | | 2 | any | -2 | | 3 | 4 | any | 6. **Explanation:** - Equation 1 is a line with slope $\frac{4}{3}$ and y-intercept $-\frac{1}{3}$. - Equation 2 is a horizontal line at $y = -2$. - Equation 3 is a vertical line at $x = 4$. These points and lines can be plotted on the Cartesian plane accordingly.