1. **Problem statement:** Solve the system where $\frac{x}{3} = \frac{y}{4}$ and $x^2 + y^2 = 100$. 2. **Set a common ratio:** Let $\frac{x}{3} = \frac{y}{4} = k$. 3. **Express variables in terms of $k$:** $$x = 3k$$ $$y = 4k$$ 4. **Substitute into the second equation:** $$x^2 + y^2 = 100$$ $$ (3k)^2 + (4k)^2 = 100$$ $$9k^2 + 16k^2 = 100$$ $$25k^2 = 100$$ 5. **Solve for $k^2$:** $$k^2 = \frac{100}{25} = 4$$ 6. **Find $k$:** $$k = \pm 2$$ 7. **Find $x$ and $y$ for each $k$:** - For $k=2$: $$x = 3 \times 2 = 6$$ $$y = 4 \times 2 = 8$$ - For $k=-2$: $$x = 3 \times (-2) = -6$$ $$y = 4 \times (-2) = -8$$ **Final answer:** $$\boxed{(x,y) = (6,8) \text{ or } (-6,-8)}$$