Quick question about free variables

  • Context: Undergrad 
  • Thread starter Thread starter wumple
  • Start date Start date
  • Tags Tags
    Variables
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
wumple
Messages
57
Reaction score
0
Hopefully I have this in the right place, it's not a homework question exactly, rather a question I have as I'm reading through my text. I'm learning about free variables. The book gives the example system:

x - z = 2
y + 2z = -1
0 = 0

as an example of a system with infinitely many solutions. I see that the way to express the solution set for this system is by describing the line that the solutions lie on. It says to make z the 'free variable' and make x and y the 'dependent' variables. Then by picking z I can find values for x and y that work. This makes sense to me. But why can't I solve for, say, x and z in terms of y? Why do I have to pick z as the free variable?
 
Physics news on Phys.org
The free parameter you choose is arbitrary, so just as you could say, like the book, that z is free and then

[tex] (x,y,z)=(2+z,-1-2z,z)[/tex]

you could also have said that x was your free parameter and stated the solutions as

[tex] (x,y,z)=(x,3-2x,-2+x)[/tex]

I think they must have chosen z because, in this case, it is easier to solve for x and y.