# 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 ?

• without hardy cross it would be difficult definitely – Aman Rajput Jun 2 at 7:12
• The general method is using Gröbner bases. – Wuestenfux Jun 2 at 7:44
• Okay let me check that.. – learningstudent Jun 2 at 7:58
• I don't understand any of this head loss business, but your system of pipes has $0.2+Q_1+Q_6$ flowing in, and $0.2$ flowing out. It would seem that $Q_1=Q_6=0$, and hence $Q_2=Q_3=0$ and $Q_4=Q_5=0.2$. This might not be consistent with the last six equations, as Quasi's calculations seem to show. – vadim123 Jun 2 at 17:52

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.