This video explains how to convert functions from slope-intercept form, and point-slope form, to the standard form of the equation of a line. After you finish this lesson, view all of our Algebra 1 lessons and practice problems.

The standard form of a line is written as:

$Ax + By = C$

where A, B, and C, are constants.

An example would be:

$y = 3x + 7$

Would be written as:

$3x - y = -7$

## Examples of Standard Form Of The Equation Of A Line

### Example 1

$2x=9-y$

The standard form of a line is written as:

$Ax + By = C$

Let’s bring all the variables to one side by adding $y$ on both sides
$2x+y=9-y+y$
Therefore, we have:
$2x+y=9$

### Example 2

$3(x+5)-2=20+y$
The standard form of a line is written as:

$Ax + By = C$

First, distribute $3$ in each of the terms inside the parenthesis.

$3x+15-2=20+y$
$3x+13=20+y$

Let’s bring all the variables to one side by adding $y$ on both sides

$3x+13-y=20+y-y$
$3x+13-y=20$

Then bring all the constant on one side by subtracting 13 on both sides

$3x+13-y-13=20-13$

Therefore, we have:

$3x-y=-7$

## Video-Lesson Transcript

The standard form of a line is

$Ax + By = C$

Where $(x-term) + (y-term) = \#$.

For example:

$y = 3x + 7$

Let’s solve to make this look like this $(x-term) + (y-term) = \#$

Let’s get all the terms on one side first by subtracting $3x$ on both sides

$y = 3x + 7$
$-3x + y = 3x - 3x + 7$
$-3x + y = 7$

It looks like it but we want the $x$ term to be positive. So we divide both sides by $-1$

$\dfrac{-3x}{-1} + \dfrac{y}{-1} = \dfrac{7}{-1}$
$3x - y = -7$

Let’s solve another one

$y - 5 = 2 (x + 7)$
$y - 5 = 2x + 14$

Let’s bring all the variables to one side by subtracting $y$ on both sides.

$- 5 = 2x - y + 4$

Then bring all the constant on one side by subtracting $4$

$- 5 - 4 = 2x - y + 4 - 4$
$- 9 = 2x - y$

Let’s just rewrite it

$2x - y = -9$

Just to recap it’s just $(x-term) + (y-term) = \#$.