How do you write $y = 3x + 1$ in standard form?

Hint: In this question, we want to write the given expression in standard form. The standard form of the linear equation of two variables is $ax + by = c$. Where ‘a’ is the coefficient of x, ‘b’ is the coefficient of y, and ‘c’ is a constant term. Here, a, b, and c are integers, and ‘a’ is non-negative. And x and y are variables. The linear equation has only the first power.

Complete step-by-step answer:
In this question, we want to write the given expression in standard form, and the given expression is:
$ \Rightarrow y = 3x + 1$
As we know the standard form of the linear equation is $ax + by = c$.
For that, subtract both sides by 3x to move x term to the left-hand side of the equation.
$ \Rightarrow - 3x + y = 3x - 3x + 1$
Let us simplify the right-hand side of an equation.
Here, subtraction of 3x and 3x is 0.
Therefore,
$ \Rightarrow - 3x + y = 1$
Now, multiply both sides of the equation by -1 to make the coefficient of x term positive.
 $ \Rightarrow \left( { - 1} \right)\left( { - 3x + y} \right) = 1\left( { - 1} \right)$
To remove the brackets multiply -1 with each term of the left-hand side of the equation.
$ \Rightarrow \left( { - 1} \right)\left( { - 3x} \right) + \left( { - 1} \right)\left( y \right) = - 1$
The multiplication of two negative numbers is positive.
$ \Rightarrow 3x - y = - 1$

Hence, this is in the standard form.

Note:
In the linear equation a, b, and c have no common factors other than 1. Standard form of an equation is useful for finding the x and y intercepts of a graph. That is, the point where the graph crosses the x-axis and the point where it crosses the y-axis. Standard form of an equation is useful for finding the points where two or more functions intersect.