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\begin{eqnarray} a_1x^2+b_1xy+c_1x+d_1z=0\\ a_2z^2+b_2zw+c_2x+d_2z=0 \\ a_3xy+b_3y+c_3w=0\\ a_4zw+b_4y+c_4w=0 \end{eqnarray} all coefficients are non-zero here.

in particular \begin{eqnarray} -10x^2-2xy+9x+2z=0\\ -10z^2-2zw+x-2z=0 \\ 100xy-10y+7w=0\\ 100zw+5y-20w=0 \end{eqnarray}

I have two questions, kindly have a look

1) Is there any algebraic or numerical method (simple or advance) exist to solve these simultaneous equations.

2) Can we conclude that this set of equation has unique solution, say, $S=(x_s,y_s,z_s,w_s)$ $\iff$ corresponding jacobian matrix (left hand side of equations represent the function of variables $x,y,z,w$) invertible at the solution point $S$ and it's neighborhood?

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    @LutzL can you give references, please. – 2017-02-11
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    Look up Bezout theorem on the number of roots of the homogenized system. I think you can explicitly compute all the solutions at infinity so that you know how many finite solutions there are. https://en.wikipedia.org/wiki/B%C3%A9zout's_theorem – 2017-02-11
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    On second thought, you have a full line/circle $(0:0:y:0:w)$ at projective infinity, which might prohibit the use of the Bezout theorem. Note that the origin is always a solution, and in general a simple one. – 2017-02-11
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    Just out of curiosity : for the specific case, there are nine solutions, all of them being real. – 2017-02-11
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    @ClaudeLeibovici did you do some algebra? – 2017-02-11
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    I am too lazy for that ! I just wanted to look at. There are a few simple solutions but all the other are quite nasty requiring the zeros of high order polynomials. – 2017-02-11
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    @ClaudeLeibovici please share your knowledge. – 2017-02-12

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