1. **Problem statement:** Find $x$ and $y$ in the two right triangles where the first triangle has sides 5 (hypotenuse), 3 (one leg), and $x$ (unknown leg), and the second triangle has sides $x$ (hypotenuse), $y$ (unknown leg), and 2 (other leg). 2. **Formula used:** We use the Pythagorean theorem for right triangles: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 3. **Find $x$:** For the first triangle, $$x^2 + 3^2 = 5^2$$ $$x^2 + 9 = 25$$ $$x^2 = 25 - 9$$ $$x^2 = 16$$ Since $x > 0$, $$x = \sqrt{16} = 4$$ 4. **Find $y$:** For the second triangle, $$y^2 + 2^2 = x^2$$ Substitute $x = 4$: $$y^2 + 4 = 16$$ $$y^2 = 16 - 4$$ $$y^2 = 12$$ $$y = \sqrt{12} = 2\sqrt{3}$$ **Final answers:** $$x = 4$$ $$y = 2\sqrt{3}$$