# Infinite number of solutions

1. Dec 31, 2012

### unscientific

1. The problem statement, all variables and given/known data

I worked out until the last part of the question and 3 equations with 3 unknowns got reduced to this:

x - 2y + 3z = 1
x + 3z = 3

3. The attempt at a solution

y = 1,
x = 3 -3z

Letting x = λ where λ is any real number,

(x,y,z) = (3,1,0) + λ(-3,0,1)

It wouldn't make a difference if i let z be λ instead right?

2. Dec 31, 2012

### HallsofIvy

Staff Emeritus
The solution you give is the same as $x= 3- 3\lambda$, $y= 1$, and $z= \lambda$. Since $z= \lambda$, it doesn't matter which you use. You can write the solution as $(x, y, z)= (3-\lambda, 1, \lambda)$ or as $(x, y, z)= (3-z, 1, z)$.