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

Hint: In this question we have to find whether the given relation defines the function or not. For this we need to be clear about the relations and functions. To check whether a relation defines a function or not, first we will separate \[y\] and express it in the form of \[x\] .after that we will check whether each \[x\] has a unique value of \[y\] or not. If each value of \[x\] has a unique value of \[y\] then the given relation defines a function, otherwise the given relation does not define a function.

Complete answer:
Let us first recall the definition of relation and function:
A relation is basically a relationship between \[x\] and \[y\] coordinates.
A function is a type of relation in which each \[x\] has a unique value of \[y\] .
Now the given relation is,
\[x - 3y = 2\]
Now, first we will separate \[y\] and express it in the form of \[x\]
So, subtracting \[ - x\] from both the sides, we get
\[x - 3y - x = 2 - x\]
\[ \Rightarrow - 3y = 2 - x\]
Taking \[ - 1\] common from the above equation, we get
\[ \Rightarrow 3y = x - 2\]
Dividing by \[3\] on both sides, we get
\[ \Rightarrow y = \dfrac{{x - 2}}{3}\]
\[ \Rightarrow y = \dfrac{1}{3}x - \dfrac{2}{3}\]
Now we see that for every \[x \in \mathbb{R}\] , \[f\left( x \right)\] maps \[x\] to a unique value of \[y\] given by \[y = \dfrac{1}{3}x - \dfrac{2}{3}\]
So, we conclude that \[x - 3y = 2\] represented by \[f\left( x \right) = y = \dfrac{1}{3}x - \dfrac{2}{3}\] is indeed a function.

Note:
One may note that if we have a relation in which we have one value of \[y\] for more than one value of \[x\] then the relation will be considered as a function. That is called a many-one function.
Also note that this question can also be done using another method. i.e.,
First of all, draw the graph of the given line. Then draw vertical lines which are parallel to the y-axis passing through the given straight line. If any one of these vertical lines passes through two points on the given straight line which have different y-coordinates then the given relation will not be a function, otherwise the given relation will be a function. Hence you will get the required result.