1. **State the problem:** We need to solve the system of linear equations by graphing and fill in the tables for each equation at $x=0$ and $x=5$. 2. **Equations given:** $$y=\frac{4}{5}x+3$$ $$y=\frac{8}{5}x-1$$ 3. **Calculate $y$ values for $y=\frac{4}{5}x+3$:** - When $x=0$: $$y=\frac{4}{5}\times0+3=0+3=3$$ - When $x=5$: $$y=\frac{4}{5}\times5+3=4+3=7$$ 4. **Calculate $y$ values for $y=\frac{8}{5}x-1$:** - When $x=0$: $$y=\frac{8}{5}\times0-1=0-1=-1$$ - When $x=5$: $$y=\frac{8}{5}\times5-1=8-1=7$$ 5. **Fill in the tables:** For $y=\frac{4}{5}x+3$: \begin{tabular}{c|c} x & y \\\hline 0 & 3 \\ 5 & 7 \\ \end{tabular} For $y=\frac{8}{5}x-1$: \begin{tabular}{c|c} x & y \\\hline 0 & -1 \\ 5 & 7 \\ \end{tabular} 6. **Interpretation:** Both lines intersect at the point where $x=5$ and $y=7$, which is the solution to the system. 7. **Final answer:** The solution to the system is $$\boxed{(5,7)}$$.