I already turned in my homework, but I do not think my solution to problem 3.34 was very good. I rewrote the first equation as $y^3 - 18xy = -x^3$ and factored it as $y(y^2 - 18x) = -x^3$. From the second equation, I rearranged it as $y^2 = 20x - x^2$ and substituted this into the first equation to get $y(2x - x^2) = -x^3$. Isolating $y$, I got $y = \frac{x^3}{x^2 - 2x}$. I substituted this into $x^2 + y^2 - 20x = 0$. After getting rid of all the denominators, I got a polynomial. By factoring, using the rational root theorem, and the quadratic formula, I found four different solutions for $x$. Substituting these values back into the original equations, I found the corresponding values of $y$. However, there were a lot of extraneous solutions for $y$, and checking for extraneous solutions took a long time since cubing $a + \sqrt{b}$ took a long time. Is there a better or faster way to solve this problem? Thank you.