# Non-linear system of equations with three unknowns

I have the following non-linear system of equations

$$\begin{cases} \tau + e^{(m+\frac{s}{2})} &= a\,\quad(1)\\ \tau^2 + 2\tau\,e^{(m+\frac{s}{2})}+e^{(2m+2s)} &= b.\quad(2)\\ \tau^3+3\tau^2\, e^{(m+\frac{s}{2})}+3\tau\,e^{(2m+2s)}+e^{(3m+\frac{9s}{2})}&= c\quad (3) \end{cases}$$

and would like to solve for m, s and $$\tau$$.

I have tried the following:

Letting $$x = e^{m}$$ and $$y=e^{s/2}$$ so as to get

$$\begin{cases} \tau + xy &= a\,\quad(3)\\ \tau^2 + 2\tau xy+ x^2y^4&= b.\quad(4)\\ \tau^3 + 3\tau^2xy+ 3\tau\,x^2y^4+x^3y^9 &= c \quad(5) \end{cases}$$

Then solving for x and y and in turns m and s.

However I'm struggling to get $$\tau$$ in terms of the unknowns only.

Any suggestions would be very much appreciated.

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa May 28 at 22:01
• To solve for m, s, and t in terms of a and b requires three equations. – William Elliot May 28 at 23:01
• My mistake, thank you I have edited my question. – joshpushy May 28 at 23:29
• "$\tau$ in terms of the unknowns only." ??? – Yves Daoust May 29 at 21:03

$$\begin{cases} t + xy &= a\,\quad(3)\\ t^2 + 2t xy+ x^2y^4 &= b\quad(4)\\ t^3 + 3t^2xy+ 3t\,x^2y^4+x^3y^9 &= c \quad(5) \end{cases}$$ $$xy=a-t\quad$$ that we put into $$(4)$$ and $$(5)$$.

$$\begin{cases} t^2 + 2t (a-t)+ (a-t)^2y^2 &= b\quad(6)\\ \color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2y^2+(a-t)^3y^6 &= c \quad(7) \end{cases}$$

from $$(6)$$ : $$y^2=\frac{b-t^2 - 2t (a-t)}{(a-t)^2}=\frac{b+t^2 - 2at}{(a-t)^2} \quad(8)$$ We put $$(8)$$ into $$(7)$$ : $$\color{red}{t^3} + 3t^2(a-t)+ 3t\,(a-t)^2\frac{b+t^2 - 2at}{3t\,(a-t)^2}+(a-t)^3\left(\frac{b+t^2 - 2at}{3t\,(a-t)^2} \right)^3=c$$ After simplification : $$\alpha t^3+\beta t^2 +\gamma t+\delta = 0\quad \begin{cases} \alpha=c-3ab+2a^3 \\ \beta =3(b^2+a^2b-a^4-ac) \\ \gamma=3a(a^2b-2b^2+ac\\ \delta=b^3-a^3c \end{cases}\tag 9$$ Then, solve the cubic equation for $$t$$.

Put $$t$$ into $$(8)$$. This gives $$y$$.

Then with $$t$$ and $$y$$ obtained, $$\quad x=\frac{a-t}{y}$$

Obviously the formulas for $$t$$, $$y$$ and $$x$$ will be very complicated.

• You may have made some typos. There should be no $3t$ in the denominator of $(8)$. (If you look at $(6)$, there's no $3t$ there.) And eliminating $y$ between $(6)$ and $(7)$ using resultants, I get a quintic in $t$, not a cubic. But this was a clever approach, nonetheless. – Tito Piezas III May 31 at 16:37
• OK. There was a typo in Eq.$(8)$ in copying. Now corrected. Fortunately the draft was not deleted yet. I re-checked it and found no mistake. So, I think that the final equation $(8)$ is correct. Thank you very much for your checking. – JJacquelin May 31 at 21:05
• I corrected another typo. You wrote $t(a-t)^2$ instead of $t^3$, which was why I ended up with a quintic in $t$. The corrected typo now yields your result that is a cubic in $t$. – Tito Piezas III Jun 1 at 2:51
• So, finally $(9)$ is correct. All's well that ends well. Cheers! – JJacquelin Jun 1 at 4:57
• Yes, all's well that ends well. :) – Tito Piezas III Jun 1 at 5:09

Let $$\tau \to t$$. In pari-gp by Resultant:

? polresultant(t^3 + 3*t^2*x*y + 3*t*x^2*y^4 + x^3*y^9 - c, t + x*y - a, x)
%6 = (t^3 - 3*a*t^2 + 3*a^2*t - a^3)*y^9 + (-3*t^3 + 6*a*t^2 - 3*a^2*t)*y^5 + (2*t^3 - 3
*a*t^2 + c)*y^3
?
? %/y^3
%7 = (t^3 - 3*a*t^2 + 3*a^2*t - a^3)*y^6 + (-3*t^3 + 6*a*t^2 - 3*a^2*t)*y^2 + (2*t^3 - 3
*a*t^2 + c)
?
? polresultant(t^2 + 2*t*x*y + x^2*y^4 - b, t + x*y - a, x)
%8 = (t^2 - 2*a*t + a^2)*y^4 + (-t^2 + 2*a*t - b)*y^2
?
? %/y^2
%9 = (t^2 - 2*a*t + a^2)*y^2 + (-t^2 + 2*a*t - b)
?
? polresultant(%7, %9, y)
%10 = (c^2 + (4*a^3 - 6*b*a)*c + (4*a^6 - 12*b*a^4 + 9*b^2*a^2))*t^12 + (-12*a*c^2 + (-4
2*a^4 + 60*b*a^2 + 6*b^2)*c + (-36*a^7 + 102*b*a^5 - 60*b^2*a^3 - 18*b^3*a))*t^11 + (66*
a^2*c^2 + (198*a^5 - 264*b*a^3 - 66*b^2*a)*c + (141*a^8 - 366*b*a^6 + 120*b^2*a^4 + 162*
b^3*a^2 + 9*b^4))*t^10 + (-220*a^3*c^2 + (-552*a^6 + 666*b*a^4 + 324*b^2*a^2 + 2*b^3)*c
+ (-314*a^9 + 708*b*a^7 + 90*b^2*a^5 - 608*b^3*a^3 - 96*b^4*a))*t^9 + (495*a^4*c^2 + (10
08*a^7 - 1044*b*a^5 - 936*b^2*a^3 - 18*b^3*a)*c + (435*a^10 - 762*b*a^8 - 825*b^2*a^6 +
1212*b^3*a^4 + 429*b^4*a^2 + 6*b^5))*t^8 + (-792*a^5*c^2 + (-1260*a^8 + 1008*b*a^6 + 176
4*b^2*a^4 + 72*b^3*a^2)*c + (-384*a^11 + 372*b*a^9 + 1632*b^2*a^7 - 1308*b^3*a^5 - 1056*
b^4*a^3 - 48*b^5*a))*t^7 + (924*a^6*c^2 + (1092*a^9 - 504*b*a^7 - 2268*b^2*a^5 - 168*b^3
*a^3)*c + (211*a^12 + 78*b*a^10 - 1665*b^2*a^8 + 568*b^3*a^6 + 1569*b^4*a^4 + 162*b^5*a^
2 + b^6))*t^6 + (-792*a^7*c^2 + (-648*a^10 - 36*b*a^8 + 2016*b^2*a^6 + 252*b^3*a^4)*c +
(-66*a^13 - 204*b*a^11 + 930*b^2*a^9 + 288*b^3*a^7 - 1434*b^4*a^5 - 300*b^5*a^3 - 6*b^6*
a))*t^5 + (495*a^8*c^2 + (252*a^11 + 234*b*a^9 - 1224*b^2*a^7 - 252*b^3*a^5)*c + (9*a^14
+ 102*b*a^12 - 240*b^2*a^10 - 492*b^3*a^8 + 771*b^4*a^6 + 330*b^5*a^4 + 15*b^6*a^2))*t^
4 + (-220*a^9*c^2 + (-58*a^12 - 156*b*a^10 + 486*b^2*a^8 + 168*b^3*a^6)*c + (-18*b*a^13
+ 238*b^3*a^9 - 204*b^4*a^7 - 216*b^5*a^5 - 20*b^6*a^3))*t^3 + (66*a^10*c^2 + (6*a^13 +
48*b*a^11 - 114*b^2*a^9 - 72*b^3*a^7)*c + (9*b^2*a^12 - 42*b^3*a^10 + 6*b^4*a^8 + 78*b^5
*a^6 + 15*b^6*a^4))*t^2 + (-12*a^11*c^2 + (-6*b*a^12 + 12*b^2*a^10 + 18*b^3*a^8)*c + (6*
b^4*a^9 - 12*b^5*a^7 - 6*b^6*a^5))*t + (a^12*c^2 - 2*b^3*a^9*c + b^6*a^6)
?
? default(parisize,16G)
***   Warning: new stack size = 16000000000 (15258.789 Mbytes).
?
? factor(%10)
%12 =
[t - a 6]

[(c + (2*a^3 - 3*b*a))*t^3 + (-3*a*c + (-3*a^4 + 3*b*a^2 + 3*b^2))*t^2 + (3*a^2*c + (3*b
*a^3 - 6*b^2*a))*t + (-a^3*c + b^3) 2]


Or in W.Mathematica. I.e. get cubic equation with unknown $$t$$ and parameters $$a,b,c$$.