Solve the equations: $$\begin{cases} 3x &- &4y &+ &1/yx &= &12\\ 4z &- &12x &+ &1/xz &= &3\\ 12y &- &3z &+ &1/yz &= &4\\ \end{cases}$$
By observation $x=1/3$, $y=1/4$, $z= 1$ is a solution. How to approach this question? I tried by multiplying third equation with second and substracting from first to make perfect squares but it's not getting solved.
Taking resultants we see that all solutions are given by $$ (x,y,z)=(1/3,1/4,1), (-1/3,-1/4,-1) $$ and $$ y=( - 456976000x^5 + 1116535680x^4 - 2394256124x^3 - 1163624319x^2 - 556971724x + 138127536)/23037439, $$
$$ z=( - 114244000x^5 + 253276920x^4 - 535731521x^3 - 452777364x^2 - 177971056x + 10655316)/26578395, $$
where $x$ is a root of the polynomial
$$ 676000x^6 - 1642680x^5 + 3555929x^4 + 1825200x^3 + 870688x^2 - 182520x + 23409=0. $$ Since this polynomial has only non-real roots, the only real solutions are the ones given above.
