Solve the following system of equations:
$\left\{ \begin{align}
& 2{{x}^{2}}-5xy-5{{y}^{2}}+x+10y-35=0 \\
& {{x}^{3}}+5x{{y}^{2}}+42=0 \\
\end{align} \right.$
By using a computer, I have solutions: $x=-3$,$y=1$, but I cannot seem to do it by hand. Please help me solve it. Thanks.
On
If you care about all solutions and not just some nice ones, especially if you are using a computer to do the grunt work, fire up any CAS (computer algebra system) at your disposal. Any self-respecting CAS has an implementation of groebner basis algorithms (to use for problems like this under the hood!) - but try sympy! The lex order groebner basis of your system is
$$52704x - 6250y^5 + 27700y^4 - 90775y^3 + 220495y^2 - 297625y + 304567=0$$ $$10y^6 - 30y^5 + 123y^4 - 246y^3 + 408y^2 - 480y + 215=0$$ The last equation is a univariate six-degree polynomial that factors into
$$(y-1)(y+i\sqrt5)(y-i\sqrt5)(y-1/2 + 9i/10\sqrt5)(y-1/2 - 9i/10\sqrt5)$$
So like your solution we have $y=1$ as well. Backsubtition into the first equation (of the groebner basis) gives $$52704x + 158112=0$$ but $-158112/52704=-3$. So indeed $y=1$ and $x=-3$ is a solution...
This fits nicely with what Dietrich posted: As he points out if $x=2$ then $y$ is a solution to $y^2=-5$ both of these solutions (the positive and negative solution of that quadratic) also solve the equation in the groebner basis that only contains $y$ (and not $x$). Which you also see from the factorization. Substituting $y=i\sqrt5$ into the first equation gives $$52704x - 105408=0$$ and we have $105408/52704=2$ so we have another nice solution of $y=i\sqrt5$ and $x=2$! The same for $y=-i\sqrt5$.
Since $x\neq 0$ by the second equation we have $y^2=-(x^3+42)/(5x)$. Substitute this to the first equation to obtain $$ y(5x)(x-2)=(3x^2 + 7x - 21)(x - 2). $$ For $x=2$ we obtain $y^2=-5$. Otherwise we can divide by $5x(x-2)$ to obtain $y$. This leads by the second euqation to $(2x^2 - 6x + 7)(x + 3)=0$, so we obtain altogether four solutions for $x$, i.e., $x=2,x=-3,x=( \pm \sqrt{-5} + 3)/2$.