1. **State the problem:** We have a table of values for $x$ and $y$ where $x$ ranges from $-3$ to $3$ and $y$ is given as $1$ when $x=0$. We need to complete the table, graph the line representing Mica's rule, and verify if the equation $2x + y = -3$ matches the graph and table. 2. **Complete the table:** Since only $y=1$ at $x=0$ is given, we need to find the pattern or rule for $y$ in terms of $x$. The point plotted is $(-3,3)$, so when $x=-3$, $y=3$. 3. **Check Mica's equation:** Mica's equation is $2x + y = -3$. Solve for $y$: $$y = -3 - 2x$$ 4. **Calculate $y$ values using Mica's equation for each $x$:** - For $x=-3$: $y = -3 - 2(-3) = -3 + 6 = 3$ - For $x=-2$: $y = -3 - 2(-2) = -3 + 4 = 1$ - For $x=-1$: $y = -3 - 2(-1) = -3 + 2 = -1$ - For $x=0$: $y = -3 - 0 = -3$ - For $x=1$: $y = -3 - 2 = -5$ - For $x=2$: $y = -3 - 4 = -7$ - For $x=3$: $y = -3 - 6 = -9$ 5. **Compare with given data:** The table shows $y=1$ at $x=0$, but Mica's equation gives $y=-3$ at $x=0$. This contradicts the given data. 6. **Find the correct equation:** Using the points $(0,1)$ and $(-3,3)$, find the slope $m$: $$m = \frac{3 - 1}{-3 - 0} = \frac{2}{-3} = -\frac{2}{3}$$ Use point-slope form with point $(0,1)$: $$y - 1 = -\frac{2}{3}(x - 0)$$ $$y = 1 - \frac{2}{3}x$$ Rewrite in standard form: $$y = 1 - \frac{2}{3}x$$ Multiply both sides by 3: $$3y = 3 - 2x$$ Rearranged: $$2x + 3y = 3$$ 7. **Verify the new equation with points:** - For $x=0$: $y=1$ correct. - For $x=-3$: $y=1 - \frac{2}{3}(-3) = 1 + 2 = 3$ correct. **Final answer:** - Mica's equation $2x + y = -3$ is incorrect. - The correct equation is $$2x + 3y = 3$$