1. Problem: Solve the simultaneous equations $3x^2-xy=0$ and $2y-5x=1$. 2. Factor the first equation to find relations between $x$ and $y$. $$3x^2-xy=x(3x-y)$$ Thus $x=0$ or $y=3x$. 3. Case 1: $x=0$, substitute into the second equation. $$2y-5(0)=1$$ $$2y=1$$ So $y=rac{1}{2}$. 4. Case 2: $y=3x$, substitute into the second equation. $$2(3x)-5x=1$$ $$6x-5x=1$$ So $x=1$. Then $y=3x=3$. 5. Final answer: The solutions are $ (0,rac{1}{2}) $ and $ (1,3) $.