Solving system of 10 equations involving some degree 2 Recently I was doing a pipe network problem without Hardy Cross Method (approximate method )
I have obtained 10 desired equations with 10 unknowns as shown in this image below highlighted by red pen.
I tried solving manually but stumped .. then i thought to use mathematical software... but except wolframalpha i dont know any other softwares
Can anyone know how to solve them... as wolframalpha feels lazy to show answer :D
How to solve using software also ?

 A: Elementary algebra is all that's needed -- tedious, but routine.

Eliminating the variables one at a time, we get
\begin{cases}
Q_2=Q_1 - {\large{\frac{K+125}{10K}}}
\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;
\\[4pt]
Q_3={\large{\frac{K+125}{10K}}}\\[4pt]
Q_4=Q_1 + {\large{\frac{K-125}{10K}}}\\[4pt]
Q_5={\large{\frac{K-125}{10K}}}\\[4pt]
Q_6=-Q_1\\[4pt]
\end{cases}
\begin{cases}
\,h_1=25-KQ_1
\\[4pt]
\,h_2=
-2KQ_1^2
+
\left({\large{\frac{K+125}{5}}}\right)Q_1
-
{\large{\frac{K^2-2250K+15625}{100K}}}
\\[4pt]
\,h_3=20-KQ_1^2\\[4pt]
\,h_4=
-KQ_1^2
-
{\large{\frac{K^2-2250K+15625}{100K}}}
\\[4pt]
\end{cases}
where $Q_1$ satisfies the quadratic equation
$$(200K^2)Q_1^2-(5000K)Q_1+(K^2-250K+15625)=0$$
But for the given value of $K$, one can verify that the discriminant of the above quadratic is negative, hence the given system of equations has no real solutions.
A: Sorry i am not able to add comment , but i think that we can make these equations into a 10×10 matrix and then we can reduce the matrix to reduced row echelon(Gauss Jordon method) form to get the solution. Then we will get values of H(i) and all Q^2 .From Q^2 we will get values of all Q's
