How to Write an Equation in Standard Form – An Explanation
As you are on this page, we assume that you are having an understanding of some basic concepts. These concepts should be clear in order to write an equation in standard form. You can write an equation in a standard format or in a slope-intercept form. But also note that you can easily convert an equation written in slope-intercept form into a standard equation form.
Following are the basic concepts used throughout in this article to write and describe the equations.
- Coordinate plane and Graphing lines
- Slope and y-intercept
- Slope-intercept and point-intercept forms
- Equations of parallel lines and perpendicular lines
- what are integersand variables
An example of a standard equation form is Ax + By = C where:
- A, B and C are integers with no common factor except 1
- Integer A is a non-negative
- x and y are variables.
Above equation can be written as y = m x + b in slope-intercept form.
Converting a slope-intercept form into a standard form
In order to change the slope-intercept into a standard equation, you have to express the coordinate of the y-intercept and slope as the quotient of two integers, which can be called a rational number form. The change in y divided by the change in x defined as slope formula. This is not a much-asked problem, but expressing the ychange/xchange, we can meet the same.
The slope-intercept form y = m x + b can also be written as y = (ychange/xchange) x + b_num/b_den. Because the ordinate of y-intercept generally follows the same system, so “b” in y = m x +b can be expressed as a quotient of two integers that are called b_num and b_den.
If we see a linear equation of a standard formula, we can observe that x and y-intercepts the corresponding line. For example, 3x + 4y = 24. Here, if we consider the value of x as 0, the resulting equation will be 4y = 24. Now we can easily tell that the value of y is 6 and therefore the y-intercept is (6, 0). In a similar way, if we consider the value of y as 0, we will get the value of x, i.e. 8 as the equation will be 3x = 24 and x-intercept will be (0, 8).
By multiplying both the sides of equation y = m x + b by the least common multiple of xchange and b_den, there will be no fractions in the resulting equation and will look like Dy = Ex + F. Here D, E and F are integers. To get this equation in a standard format, we will have to have a coefficient of x as a non-negative. So we will need to add –Ex to both the sides of the equation and will get –Ex + Dy = F. If the value of –E is negative, both the sides can be multiplied by –1.
As a result, we get an equation with a standard format, i.e. Ax + By = C, where A is a non-negative and all three, i.e. A, B and C are integers. They will not have any common factor except 1.
Examples of equations converted to standard form
To start with, we will take a complex example, as below.
y = (5/6)x + 7/4
As a first step, we will use the least common multiple of 6 and 4, i.e. 12 to multiply both the sides.
So it will be;
12y = 12( (5/6)x + 7/4 )
12y = (12)(5/6)x + (12)(7/4)
12y = 10x + 21
Then we will remove the fractions and take x to the left side of the equation.
–10x + 12y = 21
now, because of the negative coefficient of x, we need to multiply both sides by –1.
(–1)(–10x +12y) = (–1)(21)
As a result, we get a standard form of the equation as below.
10x + –12y = –21
One more good example of changing a slope-intercept equation to a standard format is as below.
y = (1/3)x + 5/6
3 and 6 have the least common multiple as 6, so we will multiply both sides by 6 to remove the fractions
6y = 6 ( (1/3)x + 5/6)
6y = (6)( (1/3)x + (6)(5/6)
6y = 2x + 5
Now add –2x to both sides, i.e.
–2x + 6y = + 5
Now multiply both the sides by –1 to make the coefficient of x positive and get a standard equation.
(–1)(–2x + 6y ) = (–1)(5)
(–1)(–2x) + (–1)(6)y ) = –5
2x + –6y = –5
Here is a simple example of the slope-intercept form
y = –2x + 8
In this equation, we don’t need to remove any fraction. Just move x term to the left side, so we get a standard equation. i.e. 2x +y = 8.
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Reasons for converting equation to standard form
We examined from that above examples that how can we convert the slope-intercept equation to a standard format. But you may have a question on why do we need to do this?
There are many reasons for this, as shown below;
- The standard equations can be written in vertical lines, but the slope-intercept form of equation cannot be written in this way.
- This form allows us to use the technique for solving linear equations.
- Finding parallel and perpendicular lines become easy and simplified with standard form calculator
Note that vertical lines can’t be used for the slope-intercept for as it has an undefined slope. But still, we can take points (4, 7) for the vertical line, i.e. 1x + 0y = 4 which can be simply written as x = 4.
Remember that a horizontal line through (4, 7) has a slope-intercept equation, i.e. y = 0x + 7 which has a standard form equation i.e. 0x + 1y = 7. This makes it clear why we need a coefficient of x as non-negative. The coefficient x should be 0 for horizontal lines.
Parallel line problem
Here is an example where we have to find the equation of a line which is parallel to 3x + 4y =17. This line contains the points (2, 8). We have to find a slope of this line and use it along with the points with the point slope form. Take a look at a standard format of the linear equation below.
Ax + By = C
and now move the first term, i.e. Ax to the other side, so it will become
By = -Ax + C
Now if we divide both the sides by B, assuming it is not 0, we have an equation in slope-intercept form, i.e.
y = (–A/B) x + C/B.
We can see that the slope is -A/B in this form. Any line that is parallel to given line should be having the same slope.
Only A and B from the slope are affecting, so as long as they are not changed, any line with a standard form of Ax + By = H will go along with the line Ax + By = C.
Now check the original problem, i.e. Make an equation of a line which is parallel to
3x + 4y =17
containing the points (2, 8). The result should look like
3x + 4y = H
where we have to find the value of H. We already know that the point (2, 8) must be making the equation true. So we will have
3(2) + 4(8) = H, that is true and therefore we have 6 + 32 = H or 38 = H
Now we know that the value of H is 38, and therefore, we can have a complete equation as
3x + 4y = 38
When we are given a standard form of the linear equation and asked to find out an equation for a parallel line using given points, we already know that the resulting equation will be looking similar to the original equation. The difference will be the constant value as the new equation will be having a different one.
To find that value, we have to substitute the initial point into the equation, for example;
2x + -5y = -19 which has the point (4, -7).
The answer of this will be 2x + -5y = H in which we will substitute the point for x and y. The new equation will be
2(4) + –5(–7) = H, therefore 8 + 35 = H and finally 43 = H
Now we have the value of H, that is 43, and the solution will be
2x + -5y = 43
From the above examples, we can say that the parallel and perpendicular line problems can be solved in a few steps. We also neither have to find the original line slope nor have to use the point-slope form.
The standard equation is 0x + 1y = C for a horizontal line. A parallel line will have a form 0x + 1y = D. So the rule will work for finding the parallel lines. For the new parallel line, we just have to get a new value for D by substituting the external point.
If we reverse the A and B values and one of its sign, for example, 0x + 1y = C to 1x + 0y = D, what we get is a vertical line according to the method of finding perpendicular lines.