1. **State the problem:** We are given a point $a(-4,1)$ and a slope $m=\frac{2}{3}$. We want to find the $y$-coordinate of point $b$ when $x=8$ on the line passing through $a$ with slope $\frac{2}{3}$. 2. **Write the point-slope form of the line:** $$y - y_1 = m(x - x_1)$$ where $(x_1,y_1) = (-4,1)$ and $m=\frac{2}{3}$. 3. **Substitute the known values:** $$y - 1 = \frac{2}{3}(x - (-4)) = \frac{2}{3}(x + 4)$$ 4. **Simplify the equation:** $$y - 1 = \frac{2}{3}x + \frac{2}{3} \times 4 = \frac{2}{3}x + \frac{8}{3}$$ 5. **Solve for $y$:** $$y = 1 + \frac{2}{3}x + \frac{8}{3} = \frac{3}{3} + \frac{2}{3}x + \frac{8}{3} = \frac{2}{3}x + \frac{11}{3}$$ 6. **Find $y$ when $x=8$:** $$y = \frac{2}{3} \times 8 + \frac{11}{3} = \frac{16}{3} + \frac{11}{3} = \frac{27}{3} = 9$$ **Final answer:** The $y$-coordinate of point $b$ when $x=8$ is $9$.