1. **State the problem:** Solve the simultaneous equations: $$3x^2 - xy = 0$$ $$2y - 5x = 1$$ 2. **From the first equation**, factor out $x$: $$x(3x - y) = 0$$ This implies two possible cases: **Case 1:** $x = 0$ **Case 2:** $3x - y = 0$ 3. **Check Case 1:** If $x=0$, substitute into the second equation: $$2y - 5(0) = 1 \\ 2y = 1 \\ y = \frac{1}{2}$$ So, $(x,y) = (0, \frac{1}{2})$ is one solution. 4. **Check Case 2:** From $3x - y = 0$, express $y$ in terms of $x$: $$y = 3x$$ Substitute this into the second equation: $$2(3x) - 5x = 1 \\ 6x - 5x = 1 \\ x = 1$$ Then find $y$: $$y = 3(1) = 3$$ So, $(x,y) = (1,3)$ is the second solution. 5. **Final answer:** The simultaneous equations have two solutions: $$\boxed{(0, \frac{1}{2}) \text{ and } (1, 3)}$$