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

Mar 25, 2016

$f \left(x\right) = y = - \frac{1}{3} x + \frac{2}{3}$ is a function since for $\forall x \in \mathbb{R}$
$f \left(x\right) : {\mathbb{R}}_{x} \to {\mathbb{R}}_{y}$ uniquely.

#### Explanation:

• Given: $x - 3 y = 2$
• 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 $f \left(x\right)$ is a function if and only if for every value of $x$ from a set $X$ $f \left(x\right)$ generates a unique mapping of $y = f \left(x\right)$ 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 $f \left(x\right) : X \to Y , \forall x \in X = \mathbb{R}$

1) First write you equation in the form: $y = f \left(x\right)$
$y = f \left(x\right) = \frac{1}{3} \left(2 - x\right) = - \frac{1}{3} x + \frac{2}{3}$
Now we see that for every $x \in \mathbb{R}$ $f \left(x\right)$ maps $x$ to a unique value of $y$, given by $y = - \frac{1}{3} x + \frac{2}{3}$.

So we conclude that $x - 3 y = 2$ represented by $f \left(x\right) = y = - \frac{1}{3} x + \frac{2}{3}$ is indeed a function.