Question

How do you decide whether the relation `x - 3y = 2` defines a function?

Answer 1

`f(x)=y=-1/3x+2/3` is a function since for `AA x in RR`
`f(x): RR_x -> RR_y` uniquely.

Explanation:

• Given: `x-3y=2`
• Required: Is the above relation a function?
• Solution Strategy: Use the standard definition of a function.
  Let X and Y be sets.
• A function `f : X → Y` means that for every `x ∈ X`, there is a unique `f(x) ∈ Y`. We call the se X the domain of `f`.

The take home lesson here is that a `f(x)` is a function if and only if for every value of `x` from a set `X` `f(x)` generates a unique mapping of `y=f(x)` in a set `Y`. i.e. `x` results in a unique single `y`.

So we need to show that the function in your question uniquely maps `f(x): X -> Y, AA x in X=RR`

1) First write you equation in the form: `y=f(x)`
`y=f(x)=1/3(2-x)= -1/3x+2/3`
Now we see that for every `x in RR` `f(x)` maps `x` to a unique value of `y`, given by `y=-1/3x+2/3`.

So we conclude that `x-3y=2` represented by `f(x)= y=-1/3x+2/3` is indeed a function.