1. **State the problem:** Solve the system of equations: $$y = -\frac{3}{2}x + 3$$ $$y = x + 8$$ Check the solution at the point $(2,16)$. 2. **Set the equations equal to each other** since both equal $y$: $$-\frac{3}{2}x + 3 = x + 8$$ 3. **Solve for $x$:** Move all terms to one side: $$-\frac{3}{2}x - x = 8 - 3$$ Simplify the right side: $$-\frac{3}{2}x - x = 5$$ Rewrite $x$ as $\frac{2}{2}x$ to combine: $$-\frac{3}{2}x - \frac{2}{2}x = 5$$ Combine like terms: $$-\frac{3+2}{2}x = 5$$ $$-\frac{5}{2}x = 5$$ Divide both sides by $-\frac{5}{2}$: $$x = \frac{5}{-\frac{5}{2}}$$ Show cancellation: $$x = 5 \times \cancel{\frac{1}{-\frac{5}{2}}} = 5 \times -\frac{2}{5} = -2$$ 4. **Find $y$ by substituting $x = -2$ into one of the original equations, e.g., $y = x + 8$:** $$y = -2 + 8 = 6$$ 5. **Solution to the system:** $$(x,y) = (-2, 6)$$ 6. **Check the given point $(2,16)$:** Substitute $x=2$ into both equations: For $y = -\frac{3}{2}x + 3$: $$y = -\frac{3}{2} \times 2 + 3 = -3 + 3 = 0$$ For $y = x + 8$: $$y = 2 + 8 = 10$$ Since $y$ values are not equal (0 vs 10), $(2,16)$ is **not** a solution to the system. **Final answer:** The solution to the system is $(-2, 6)$, and the point $(2,16)$ is not a solution.