1. **Stating the problem:** Solve the simultaneous equations $$3x^3x^2 - xy = 0$$ and $$2y - 5x = 1$$ 2. **Simplify the first equation:** Notice that $3x^3x^2 = 3x^{3+2} = 3x^5$. So the first equation becomes $$3x^5 - xy = 0$$ 3. **Factor $y$ in the first equation:** $$3x^5 = xy \implies y = \frac{3x^5}{x}$$ Since $x \neq 0$ to avoid division by zero, simplify: $$y = 3x^4$$ 4. **Substitute $y = 3x^4$ into the second equation:** $$2(3x^4) - 5x = 1 \, \Rightarrow 6x^4 - 5x = 1$$ 5. **Rewrite as a polynomial equation:** $$6x^4 - 5x - 1 = 0$$ 6. **Solve for $x$:** This quartic equation is not easily factorable by inspection. Check possible rational roots using the Rational Root Theorem (possible roots $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}$). Testing these: - For $x=1$: $6(1)^4 - 5(1) - 1 = 6 - 5 - 1 = 0$, so $x=1$ is a root. 7. **Divide the polynomial by $(x-1)$ to factor out the root:** Using polynomial division or synthetic division, we get: $$6x^4 - 5x - 1 = (x - 1)(6x^3 + 6x^2 + 6x + 1)$$ 8. **Find roots of the cubic:** $$6x^3 + 6x^2 + 6x + 1 = 0$$ Solving cubic exactly is complex; focus on the real root(s). By testing $x=-\frac{1}{2}$: $$6(-\frac{1}{2})^3 + 6(-\frac{1}{2})^2 + 6(-\frac{1}{2}) + 1 = 6(-\frac{1}{8}) + 6(\frac{1}{4}) - 3 + 1 = -\frac{3}{4} + \frac{3}{2} - 3 + 1 = -\frac{1}{4} \neq 0$$ No obvious rational root. Use numerical methods or approximate. 9. **Summary of roots:** - $x=1$ exact solution. - The cubic $6x^3 + 6x^2 + 6x + 1=0$ has one real root approximately $x \approx -0.196$ estimated numerically. 10. **Find corresponding $y$ values using $y = 3x^4$:** - For $x=1$, $$y = 3(1)^4 = 3$$ - For $x \approx -0.196$, $$y = 3(-0.196)^4 = 3(0.00147) \approx 0.0044$$ **Final solutions:** $$\boxed{(x,y) = (1,3) \quad \text{and} \quad ( -0.196, 0.0044 )}$$