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I need to find an explicit solution of this system of polynomial equations of degree $2$ in two variables $x,\,y$: $$\begin{cases} p_1x^2+q_1y^2+r_1xy+s_1x+t_1y+u_1=0\\ p_2x^2+q_2y^2+r_2xy+s_2x+t_2y+u_2=0 \end{cases}$$ Here is my approach: Solve the first equation as a quadratic equation with respect to $x$ (treating $y$ as an additional parameter) and substitute each of its two solutions $x_1(y),\,x_2(y)$ (containing radicals) into the second equation, thus eliminating $x$ from it. Then, for each substitution $x_i(y)$, it is possible (I guess) to reduce the second equation to a quartic(?) equation in one variable $y$ (possibly introducing false roots), solve it, then substitute each of the four roots $y_{i,1},\,y_{i,2},\,y_{i,3},\,y_{i,4}$ back into the corresponding solution $x_i(y)$ of the first equation, thus obtaining corresponding roots $x_{i,1},\,x_{i,2},\,x_{i,3},\,x_{i,4}$. So, in total, we will get $8$ pairs of roots $(x_{i,j},\,y_{i,j}),\,i=1..2,\,j=1..4$, that finally need to be validated against both equations to eliminate possible false roots. I tried follow this plan, but got lost in long computations.

Hopefully, this system of equations is common enough, and a simplified final solution is already known.

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    $\begingroup$ I doubt that you are going to get any really easy solution to this, but here is a slight simplification. Take $p_2$ times your first equation minus $p_1$ times the second. This will eliminate $x^2$ and you can solve for $x$ in terms of $y$: you will have only one solution and no radicals. Substitute back into either initial equation to get a quartic in $y$. $\endgroup$ – David Jun 6 '16 at 1:39
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The resultant of these two polynomials with respect to $y$ is

$$ \left( {p_{{1}}}^{2}{q_{{2}}}^{2}-2\,p_{{1}}p_{{2}}q_{{1}}q_{{2}}+p_{ {1}}q_{{1}}{r_{{2}}}^{2}-p_{{1}}q_{{2}}r_{{1}}r_{{2}}+{p_{{2}}}^{2}{q_ {{1}}}^{2}-p_{{2}}q_{{1}}r_{{1}}r_{{2}}+p_{{2}}q_{{2}}{r_{{1}}}^{2} \right) {x}^{4}- \left( 2\,p_{{1}}q_{{1}}q_{{2}}s_{{2}}-2\,p_{{1}}q_{ {1}}r_{{2}}t_{{2}}-2\,p_{{1}}{q_{{2}}}^{2}s_{{1}}+p_{{1}}q_{{2}}r_{{1} }t_{{2}}+p_{{1}}q_{{2}}r_{{2}}t_{{1}}-2\,p_{{2}}{q_{{1}}}^{2}s_{{2}}+2 \,p_{{2}}q_{{1}}q_{{2}}s_{{1}}+p_{{2}}q_{{1}}r_{{1}}t_{{2}}+p_{{2}}q_{ {1}}r_{{2}}t_{{1}}-2\,p_{{2}}q_{{2}}r_{{1}}t_{{1}}+q_{{1}}r_{{1}}r_{{2 }}s_{{2}}-q_{{1}}{r_{{2}}}^{2}s_{{1}}-q_{{2}}{r_{{1}}}^{2}s_{{2}}+q_{{ 2}}r_{{1}}r_{{2}}s_{{1}} \right) {x}^{3}- \left( 2\,p_{{1}}q_{{1}}q_{{ 2}}u_{{2}}-p_{{1}}q_{{1}}{t_{{2}}}^{2}-2\,p_{{1}}{q_{{2}}}^{2}u_{{1}}+ p_{{1}}q_{{2}}t_{{1}}t_{{2}}-2\,p_{{2}}{q_{{1}}}^{2}u_{{2}}+2\,p_{{2}} q_{{1}}q_{{2}}u_{{1}}+p_{{2}}q_{{1}}t_{{1}}t_{{2}}-p_{{2}}q_{{2}}{t_{{ 1}}}^{2}-{q_{{1}}}^{2}{s_{{2}}}^{2}+2\,q_{{1}}q_{{2}}s_{{1}}s_{{2}}+q_ {{1}}r_{{1}}r_{{2}}u_{{2}}+q_{{1}}r_{{1}}s_{{2}}t_{{2}}-q_{{1}}{r_{{2} }}^{2}u_{{1}}-2\,q_{{1}}r_{{2}}s_{{1}}t_{{2}}+q_{{1}}r_{{2}}s_{{2}}t_{ {1}}-{q_{{2}}}^{2}{s_{{1}}}^{2}-q_{{2}}{r_{{1}}}^{2}u_{{2}}+q_{{2}}r_{ {1}}r_{{2}}u_{{1}}+q_{{2}}r_{{1}}s_{{1}}t_{{2}}-2\,q_{{2}}r_{{1}}s_{{2 }}t_{{1}}+q_{{2}}r_{{2}}s_{{1}}t_{{1}} \right) {x}^{2}+ \left( 2\,{q_{ {1}}}^{2}s_{{2}}u_{{2}}-2\,q_{{1}}q_{{2}}s_{{1}}u_{{2}}-2\,q_{{1}}q_{{ 2}}s_{{2}}u_{{1}}-q_{{1}}r_{{1}}t_{{2}}u_{{2}}-q_{{1}}r_{{2}}t_{{1}}u_ {{2}}+2\,q_{{1}}r_{{2}}t_{{2}}u_{{1}}+q_{{1}}s_{{1}}{t_{{2}}}^{2}-q_{{ 1}}s_{{2}}t_{{1}}t_{{2}}+2\,{q_{{2}}}^{2}s_{{1}}u_{{1}}+2\,q_{{2}}r_{{ 1}}t_{{1}}u_{{2}}-q_{{2}}r_{{1}}t_{{2}}u_{{1}}-q_{{2}}r_{{2}}t_{{1}}u_ {{1}}-q_{{2}}s_{{1}}t_{{1}}t_{{2}}+q_{{2}}s_{{2}}{t_{{1}}}^{2} \right) x+{q_{{1}}}^{2}{u_{{2}}}^{2}-2\,q_{{1}}q_{{2}}u_{{1}}u_{{2}}- q_{{1}}t_{{1}}t_{{2}}u_{{2}}+q_{{1}}{t_{{2}}}^{2}u_{{1}}+{q_{{2}}}^{2} {u_{{1}}}^{2}+q_{{2}}{t_{{1}}}^{2}u_{{2}}-q_{{2}}t_{{1}}t_{{2}}u_{{1}} $$ Do you really want to plug that, in all its glory, in to a formula for solving a quartic? You'll get an enormous mess that is very unlikely to simplify to any great extent.

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To eliminate $y\,$ from the system, resulting in a single (quartic) equation in $x\,$, use Sylvester's dialytic eliminant, as follows:

$\begin{vmatrix} \,q_1\quad&r_1x+t_1\quad&p_1x^2+s_1x+u_1&0\\ 0&q_1&r_1x+t_1&p_1x^2+s_1x+u_1\\ q_2&r_2x+t_2&p_2x^2+s_2x+u_2&0\\ 0&q_2&r_2x+t_2&p_2x^2+s_2x+u_2 \end{vmatrix}=0$

This will give you your quartic in $x\,$ directly.

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