1. **Problem a:** Find the point of intersection of the lines $y=2x-1$ and $y=-x+8$. 2. To find the intersection, set the two expressions for $y$ equal: $$2x - 1 = -x + 8$$ 3. Solve for $x$: $$2x + x = 8 + 1$$ $$3x = 9$$ $$x = \frac{\cancel{3}x}{\cancel{3}} = \frac{9}{3} = 3$$ 4. Substitute $x=3$ into one of the original equations to find $y$: $$y = 2(3) - 1 = 6 - 1 = 5$$ 5. So, the point of intersection is $(3, 5)$. --- 6. **Problem b:** Find the point of intersection of the lines $y=\frac{1}{3}x + 2$ and $y = x - 4$. 7. Set the two expressions for $y$ equal: $$\frac{1}{3}x + 2 = x - 4$$ 8. Move terms to one side: $$\frac{1}{3}x - x = -4 - 2$$ $$\frac{1}{3}x - \frac{3}{3}x = -6$$ $$-\frac{2}{3}x = -6$$ 9. Solve for $x$: $$x = \frac{-6}{-\frac{2}{3}} = -6 \times \frac{3}{-2} = \frac{-18}{-2} = 9$$ 10. Substitute $x=9$ into one of the original equations to find $y$: $$y = 9 - 4 = 5$$ 11. So, the point of intersection is $(9, 5)$.