1. State the problem: Solve the system of equations below. $$3x^5 - x y = 0$$ $$2y - 5x = 1$$ 2. Factor the first equation to relate $x$ and $y$. From $$3x^5 - x y = 0$$ factor out $x$ to obtain $x(3x^4 - y) = 0$. This gives two cases: $x = 0$ or $y = 3x^4$. 3. Case 1: $x = 0$. Substitute $x = 0$ into $2y - 5x = 1$ to get $2y = 1$ and hence $y = 1/2$. So one solution is $(0, 1/2)$. 4. Case 2: $y = 3x^4$. Substitute into $2y - 5x = 1$ to get $2(3x^4) - 5x = 1$ which simplifies to $6x^4 - 5x - 1 = 0$. 5. Solve $6x^4 - 5x - 1 = 0$. Test $x = 1$ and find $6(1)^4 - 5(1) - 1 = 0$, so $x = 1$ is a root. Divide $6x^4 - 5x - 1$ by $x - 1$ to get $(x - 1)(6x^3 + 6x^2 + 6x + 1) = 0$. 6. Solve $6x^3 + 6x^2 + 6x + 1 = 0$ for real roots. The derivative $18x^2 + 12x + 6$ is always positive, so the cubic is strictly increasing and has exactly one real root. Using Newton's method (or a numerical solver) gives the real root approximately $x\approx -0.1981506$. The corresponding $y$ is $y = 3x^4 \approx 0.00462498$. 7. Collect the real solutions. The real solutions of the original system are $(0, 1/2)$, $(1, 3)$, and $(-0.1981506, 0.00462498)$. Final answer: $(0, 1/2), (1, 3), (-0.1981506, 0.00462498)$.