1. State the problem: Solve the simultaneous equations $3x^2 - xy = 0$ and $2y - 5x = 1$. 2. Isolate $y$ from the linear equation. $$2y - 5x = 1$$ $$2y = 5x + 1$$ $$y = \frac{5x + 1}{2}$$ 3. Substitute the expression for $y$ into the quadratic equation $3x^2 - xy = 0$. $$3x^2 - x\left(\frac{5x + 1}{2}\right) = 0$$ Multiply both sides by 2 to clear the denominator. $$6x^2 - x(5x + 1) = 0$$ Expand and simplify. $$6x^2 - 5x^2 - x = 0$$ $$x^2 - x = 0$$ Factor the left-hand side. $$x(x - 1) = 0$$ Hence $x = 0$ or $x = 1$. 4. Substitute each $x$ value back to find $y$. If $x = 0$ then $$y = \frac{5\cdot 0 + 1}{2} = \frac{1}{2}$$ If $x = 1$ then $$y = \frac{5\cdot 1 + 1}{2} = \frac{6}{2} = 3$$ 5. Final answer: The solutions are $$ (x,y) = (0, \tfrac{1}{2}),\; (1, 3) $$