System with more variables than equations!

Question

Does the system of equations have a solution? :

$a=3d$

$3b=3a+d+9e$

$3c=3b+e+9f$

$f+3c=0$

I was told by someone online that they solved it in terms of the variable $d$, in other words they got all the other variables as a function of $d$. This seems impossible because the system has more variables than equations. It has some symmetry though.

Answer 1

Noticing the first and fourth equations make for a couple of substitutions:

$3b=3(3d)+d+9e \iff 3b-10d-9e=0 \iff 3b=10d+9e$

$-f=3b+e+9f \iff 3b+e+10f=0 \iff 3b=-e-10f$

$10d+9e=-e-10f \iff 10d+10e+10f=0 \iff d+e+f=0 \iff e = -d-f$

Thus, allowing $f,d$ to be free, the solution can be expressed this way using the equations and above expressions:

$a=3d$

$b=\frac{10d+9e}{3}$

$c=-\frac{f}{3}$

$e=-d-f$

Or to use a 6-tuple: $(a,b,c,d,e,f) = (3d,\frac{d-9f}{3} , -\frac{f}{3},d,-d-f ,f) = (9x,x-9y,-y,3x,-3x-3y,3y)$ where x and y are any real numbers for a parametric solution.

Note that if x=y=1, the solution of (9,-9,-1,3,-6,3) works just as a check.

Answer 2

It does not have a unique solution, since, as you note, it has 6 unknowns and 4 equations. If every variable can be expressed as a function of $d$, we still have that $d$ can can take on any value in the domain. There would be as many solutions as there are values in the domain: infinitely many. Once $d$ would be determined, all other variables can then be determined.

Clearly, though, the solutions to the system of equations will depend on two free variables, not just $d$, since there are $6$ variables and $4$ equations. That is, four of the variables will be functions of one or both free variables.