How to solve these equations for $a$, $b$, $c$ and $x$?
I have the following:
\begin{align} 1 &= 2a+b+c\\ a &= (a+b)x + 0.25(a+c)\\ a&=(a+c)(1-x)\\ b&=a(1-x)+c(x-0.25)\\ c&=b(1-x)+a(x-0.25) \end{align}
I tried, but ended circular at the point I started. Can someone help me?
On
Using Grobner basis in maxima gives:
grobner_basis([2*a+b+c-1,(a+b)x+(a+c)/4-a,(a+c)(1-x)-a,a*(1-x)+c*(x-1/4)-b,b*(1-x)+a*(x-1/4)-c]);
(- 44) x + (- 144) a^2 + 213 a + (- 36),
(- 11) c + (- 48) a^2 + 60 a + (- 12),
(- 11) b + 48 a^2 + (- 82) a + 23,
(- 144) a^3 + 381 a^2 + (- 312) a + 64
The last equation is a cubic in "a" alone.
The second last solves for b given a.
The third last solves for c in a.
The top equation solves for x in a.
Grobner Basis and Buchberger's algorithm explain how this works.
regards arthur
lower case x
grobner_basis([2*A+B+C-1,(A+B)x+(A+C)/4-A,(A+C)(1-x)-A,A*(1-x)+C*(x-1/4)-B,B*(1-x)+A*(x-1/4)-C]);
(- 3) C + 4 x^2 ,
(- 3) B + 4 x^2 + (- 8) x + 3,
(- 3) A + (- 4) x^2 + 4 x,
(- 8) x^3 + 13 x^2 + (- 12) x + 3
If we simplify the second equation we get $$(a+b)x=\frac34a-\frac{c}4.\tag{A}$$ If we simplify the third equation, we get $$(a+c)x=c.\tag{B}$$ Now we can simply add these two equations together and use $(2a+b+c)x=x$ on the LHS to get $$x=\frac{3a+3c}4.\tag{C}$$ If we multiply this equation by $x$ again, we get $$x^2=\frac34(a+c)x\overset{(B)}=\frac34c,$$ i.e., $$c=\frac43x^2$$ So we managed to express $c$ in terms of $x$.
From (C) we have $a+c=\frac43x$. Using this we can get $a=\frac43x-c$, i.e., $$a=\frac43x-\frac43x^2=\frac43x(1-x).$$
From $2a+b+c=1$ we have $b=1-2a-c$ which leads to $$b=1-\frac83x+\frac43x^2.$$
It is relatively easy to check (simply by plugging the values into the equations) that these equations fulfill $2a+b+c=1$ and also (A) and (B). This means that the first three equations are fulfilled for any $x$.
Now it remains to check the fourth and the fifth equation. Unless I have made a mistake when transcribing the equations to WolframAlpha, both equations lead to the cubic equation $$8x^3-13x^2+12x-3=0.$$ See here and here.
This equation seems to have no rational roots. (At least Wolfram Alpha did not return any. This can be checked manually using rational roots test, although this can be a lot of work.)
So the best way is probably to try it solve numerically. There are also exact methods to find roots of cubic equations, but they tend to be too laborious to be done by hand, unless the coefficients and roots are "nice".