Solving system of 10 equations involving some degree 2

Question

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 ?

Answer 1

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.

Answer 2

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