# System with more variables than equations!

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.

• Have you considered re-writing this system in a systematic way and doing a Matrix reduction as a way to resolve it? – JB King Feb 4 '15 at 18:41
• @JBKing: Hello, I only JUST started linear algebra, I don't know most of the stuff. Do you think it's do-able by hand? – user45220 Feb 4 '15 at 18:42
• Yes, I believe it is. – JB King Feb 4 '15 at 18:42
• This can be rewritten as a homogeneous system so it will always have at least the trivial solution. This would be a more interesting question if you are looking for all the solutions rather than the existence of one. – JessicaK Feb 4 '15 at 18:45
• @JessicaK: Hello, yes I am also looking for the solutions (in terms of some variable such as $d$). Can you help? Thanks! – user45220 Feb 4 '15 at 18:47

## 2 Answers

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.

• This is the best we could do (everything in terms of 2 variables), brilliant! – user45220 Feb 4 '15 at 19:13

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.

• Thanks! But how do you solve it in terms of $d$? I've been trying for about an hour before I decided to post it here. – user45220 Feb 4 '15 at 18:44
• Actually there are 6 unknown variables and 4 equations,so they can't be solved as a function of d.You need two variables to express the other variables I think. – Zoe Feb 4 '15 at 18:51
• @user45220 Your online friend must be mistaken: there isn't any way to express all 6 variables as function of one variable. So don't waste any more time trying to do so! – Namaste Feb 4 '15 at 18:57
• @amWhy: That makes more sense. I think he must have let one of the variables =0 or something. – user45220 Feb 4 '15 at 19:12