1. The problem is to analyze and graph the system of linear equations: $$y=\frac{7}{2}x - 3$$ $$y=-\frac{3}{2}x + 7$$ 2. These are two lines in slope-intercept form $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept. 3. To find the point of intersection, set the two equations equal: $$\frac{7}{2}x - 3 = -\frac{3}{2}x + 7$$ 4. Add $\frac{3}{2}x$ to both sides: $$\frac{7}{2}x + \frac{3}{2}x - 3 = 7$$ 5. Combine like terms: $$\left(\frac{7}{2} + \frac{3}{2}\right)x - 3 = 7$$ $$\frac{10}{2}x - 3 = 7$$ $$5x - 3 = 7$$ 6. Add 3 to both sides: $$5x - \cancel{3} + \cancel{3} = 7 + 3$$ $$5x = 10$$ 7. Divide both sides by 5: $$\frac{\cancel{5}x}{\cancel{5}} = \frac{10}{5}$$ $$x = 2$$ 8. Substitute $x=2$ into the first equation to find $y$: $$y = \frac{7}{2} \times 2 - 3 = 7 - 3 = 4$$ 9. The lines intersect at the point $(2,4)$. 10. The graph shows two lines crossing at $(2,4)$ with slopes $\frac{7}{2}$ and $-\frac{3}{2}$ respectively. Final answer: The lines intersect at the point $\boxed{(2,4)}$.