In order to see how to write equations and graph in standard form, let's take an equation in the familiar slope-intercept form and convert to standard form:

y = \dfrac{2}{3}x - 1

Since the coefficients in standard form are integers, we will start by multiplying each term in the equation by 3 to eliminate the denominator in the slope.

3 \cdot y = 3\left(\dfrac{2}{3}x -1\right)

3y = 2x - 3

We will next move the x term to the other side of the equation.

-2x + 3y = - 3

Optionally, to make A positive, multiply the equation by -1.

2x - 3y = 3

To graph from this standard form, we can find and plot the x- and y-intercepts, and the resulting line will also contain the key features seen in the original slope-intercept form.

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x-intercept: substitute 0 for y and solve:

2x - 3\left(0\right) = 3

2x = 3

y-intercept: substitute 0 for x and solve:

2\left(0\right) - 3y = 3

-3y = 3

Graph the intercepts, in this case \left(\dfrac{3}{2},\,0\right) and \left(0,\,-1\right) and draw the line through the points.

We see that the graph still maintains the same slope and y-intercept as the original slope-intercept form, but standard form allows us to focus on other key features of the function.

Examples

Example 1

Consider the linear equation:\dfrac{1}{2}x+4y=16

Determine whether the equation is written in standard form. If not, rewrite it in standard form.

Worked Solution

Create a strategy

In order for the equation to be written in standard form, we must confirm that the equation is written in the form Ax+By=C, that A,\,B and C are integers, and A and B are not both zero.

Apply the idea

Since A=\dfrac{1}{2} in the equation, it's not written in standard form because A must be an integer and integers are the set of positive and negative whole numbers.

To convert the equation to standard form, we can use properties of equality and multiply each term by 2 to change the rational coefficient of x.

\displaystyle \dfrac{1}{2}x + 4y	\displaystyle =	\displaystyle 16	Original equation
\displaystyle 2 \cdot \left(\dfrac{1}{2}x + 4y\right)	\displaystyle =	\displaystyle 2 \cdot 16	Multiplication property of equality
\displaystyle x +8y	\displaystyle =	\displaystyle 32	Evaluate the multiplication

A=1, B=8, and C=32 now meet the requirements for standard form.

Graph the equation using the standard form from part (a).

Worked Solution

Create a strategy

To graph the equation, we will find the x-intercept and y-intercept and then draw the line passing through these points.

Apply the idea

To find the x-intercept, set y = 0 and solve for x:

\displaystyle x + 8(0)	\displaystyle =	\displaystyle 32	Substitute y = 0
\displaystyle x	\displaystyle =	\displaystyle 32	Simplify to find the x-intercept

So, the x-intercept is (32,\,0).

To find the y-intercept, set x = 0 and solve for y:

\displaystyle 0 + 8y	\displaystyle =	\displaystyle 32	Substitute x = 0
\displaystyle 8y	\displaystyle =	\displaystyle 32	Simplify
\displaystyle y	\displaystyle =	\displaystyle 4	Divide by 8 to find the y-intercept

So, the y-intercept is (0,\,4).

Reflect and check

By using the intercepts, we accurately graphed the equation. The line passes through both intercepts (32,\,0) and (0,\,4).

Example 2

A line with a slope of -\dfrac{1}{4} passes through the point \left(-8,5\right).

Write the equation of the line in slope-intercept form.

Worked Solution

Create a strategy

Use the slope and the point to solve for the y-intercept and write in slope-intercept form.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle mx+b	Slope-intercept form
\displaystyle 5	\displaystyle =	\displaystyle -\dfrac{1}{4}\left(-8\right)+b	Substitute y=5,\,m=-\dfrac{1}{4},\, and x=-8
\displaystyle 5	\displaystyle =	\displaystyle 2+b	Evaluate the multiplication
\displaystyle 3	\displaystyle =	\displaystyle b	Simplify
\displaystyle y	\displaystyle =	\displaystyle -\dfrac{1}{4}x + 3	Write in slope-intercept form

Convert this equation to standard form.

Worked Solution

Create a strategy

Rearrange the terms and multiply coefficients as needed to fit standard form.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle -\dfrac{1}{4}x+3	Slope-intercept form from part a
\displaystyle 4 \cdot y	\displaystyle =	\displaystyle 4\left(-\dfrac{1}{4}x + 3\right)	Multiply both sides by 4
\displaystyle 4y	\displaystyle =	\displaystyle -x+12	Evaluate
\displaystyle x + 4y	\displaystyle =	\displaystyle 12	Move the x term to the other side of the equation.

The standard form of the equation is x + 4y = 12

Solve for the x-intercept and y-intercept of the standard form equation.

Worked Solution

Create a strategy

Substitute 0 and solve to find the x-intercept,\left(x,0\right), and the y-intercept, \left(0,y\right).

Apply the idea

Finding the x-intercept:

\displaystyle x+4y	\displaystyle =	\displaystyle 12	Standard form equation from part b
\displaystyle x+4\left(0\right)	\displaystyle =	\displaystyle 12	Substitute y=0
\displaystyle x+0	\displaystyle =	\displaystyle 12	Evaluate the multiplication
\displaystyle x	\displaystyle =	\displaystyle 12	Simplify

The x-intercept is \left(12,0\right).

Finding the y-intercept

\displaystyle x+4y	\displaystyle =	\displaystyle 12	Standard form equation from part b
\displaystyle \left(0\right)+4y	\displaystyle =	\displaystyle 12	Substitute x=0
\displaystyle 4y	\displaystyle =	\displaystyle 12	Evaluate the addition
\displaystyle y	\displaystyle =	\displaystyle 3	Simplify

The y-intercept is \left(0,\,3\right).

Example 3

Consider the line shown in the graph below.

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Write the equation of the line in standard form.

Worked Solution

Create a strategy

To write the equation of the line in standard form, we will first write it in slope-intercept form y = mx + b, then convert it to standard form Ax + By = C where A,\,B, and C are integers.

Apply the idea

First, we find the slope (m) of the line.

\displaystyle m	\displaystyle =	\displaystyle \dfrac{y_2 - y_1}{x_2 - x_1}	Write the formula
	\displaystyle =	\displaystyle \dfrac{4 - 1}{0 - 2}	Substitute y_{2}=4,\,y_{1}=1,\,x_{2}=0 and x_{1}=2
	\displaystyle =	\displaystyle \dfrac{3}{-2}	Evaluate the subtraction
	\displaystyle =	\displaystyle -\dfrac{3}{2}	Simplify

The slope of the line is -\dfrac{3}{2}.

Now, we use the slope-intercept form with the point (0,\,4) to find the y-intercept (b):

\displaystyle y	\displaystyle =	\displaystyle mx+b	Write the slope-intercept form
\displaystyle y	\displaystyle =	\displaystyle -\dfrac{3}{2}x + 4	Substitute m=-\dfrac{3}{4} and b=4

Now, we convert the slope-intercept form to standard form:

\displaystyle y	\displaystyle =	\displaystyle -\dfrac{3}{2}x + 4	Original slope-intercept form
\displaystyle y \cdot 2	\displaystyle =	\displaystyle \left(-\dfrac{3}{2}x + 4\right)\cdot 2	Multiply both sides by 2
\displaystyle 2y	\displaystyle =	\displaystyle -3x + 8	Evaluate the multiplication
\displaystyle 2y+3x	\displaystyle =	\displaystyle -3x+3x+8	Add 3x to both sides
\displaystyle 3x + 2y	\displaystyle =	\displaystyle 8	Evaluate and rearrange to standard form

The equation of the line in standard form is 3x + 2y = 8.

Reflect and check

To confirm our equation, we can check the intercepts:

For the x-intercept, set y = 0:

\displaystyle 3x + 2(0)	\displaystyle =	\displaystyle 8	Substitute y = 0
\displaystyle 3x	\displaystyle =	\displaystyle 8	Evaluate the multiplication
\displaystyle \dfrac{3x}{3}	\displaystyle =	\displaystyle \dfrac{8}{2}	Divide both sides by 3
\displaystyle x	\displaystyle =	\displaystyle \dfrac{8}{3}	Evaluate

The x-intercept is \left(\dfrac{8}{3},\,0\right).

For the y-intercept, set x = 0:

\displaystyle 3(0) + 2y	\displaystyle =	\displaystyle 8	Substitute x = 0
\displaystyle 2y	\displaystyle =	\displaystyle 8	Evaluate the multiplication
\displaystyle \dfrac{2y}{2}	\displaystyle =	\displaystyle \dfrac{8}{2}	Divide both sides by 2
\displaystyle y	\displaystyle =	\displaystyle 4	Evaluate

The y-intercept is \left(0,\,4\right).

These intercepts match the points given in the original problem, confirming that the equation 3x + 2y = 8 is correct.

Example 4

A tour company travels to the Great Smoky Mountains National Park. They use a combination of buses and vans to get tourists to their destination. One bus can take 42 passengers, and one van can take 7 passengers. One day, they have 168 people register for the tour.

Write an equation in standard form that could be used to model the number of buses and vans they could use to transport all the people registered, assuming that each vehicle will be filled.

Worked Solution

Create a strategy

In words, we can start with the idea that: \left(\text{Number of people on buses}\right)+\left(\text{Number of people on vans}\right)=\text{Total number of people} and that: \text{Number of people on buses}=42 \cdot \left(\text{Number of buses}\right) and:\text{Number of people on vans}=7 \cdot \left(\text{Number of vans}\right)

We will then need to define variables and write an equation using them.

Apply the idea

Let b represent the number of buses used and let v represent the number of vans used.

So the \text{Number of people on buses}=42b and \text{Number of people on vans}=7v

which finally gives us the whole equation: 42b+7v=168

Reflect and check

It is important to declare variables and it can be helpful to use variables that relate to the quantities in the scenario to make sure we don't mix them up.

Graph the equation with an appropriate scale and labels.

Worked Solution

Create a strategy

In this case, there isn't a clear independent and dependent variable, so we can put b on the horizontal axis and v on the vertical axis.

To determine an appropriate scale, we can first find the values of the intercepts as those can give an idea of the maximum values for each axis.

Apply the idea

Find the b-intercept:

\displaystyle 42b+7v	\displaystyle =	\displaystyle 168	Equation from part (a)
\displaystyle 42b+7\left(0\right)	\displaystyle =	\displaystyle 168	Substitute v=0
\displaystyle 42b	\displaystyle =	\displaystyle 168	Evaluate the product
\displaystyle b	\displaystyle =	\displaystyle 4	Division property of equality

Find the v-intercept:

\displaystyle 42b+7v	\displaystyle =	\displaystyle 168	Equation from part (a)
\displaystyle 42\left(0\right)+7v	\displaystyle =	\displaystyle 168	Substitute b=0
\displaystyle 7v	\displaystyle =	\displaystyle 168	Evaluate the product
\displaystyle v	\displaystyle =	\displaystyle 24	Division property of equality

\text{Number of buses }\left(b\right)

\text{Number of vans }\left(v\right)

An appropriate scale would be going up by 1 along the b-axis to a maximum of 5 and going up by 2 or 4 along the v-axis to a maximum of 26. This will extend both of our axes just past where we need to plot the intercepts.

Reflect and check

Notice that it would not make sense to connect these dots with a line. This is because the situation is discrete, not continuous.

If we were to include all points that fall on the line, this would mean we could use partial buses or vans to transport people. While the vans or buses could only be partially filled, we would still require a whole number of buses and vans.

Predict the number of vans that would be required if only 1 bus was available.

Worked Solution

Create a strategy

We can find the point along the b-axis where b=1 and then up to the line and across to the v-axis to find the corresponding value for v, the number of vans.

Apply the idea

\text{Number of buses }\left(b\right)

\text{Number of vans }\left(v\right)

Using the graph, we can see that the point \left(1,\,18\right) lies on the graph of the equation. If only 1 bus was available, they would need 18 vans.

Reflect and check

We can check using the equation.

\displaystyle 42b+7v	\displaystyle =	\displaystyle 168	Equation from part (a)
\displaystyle 42\left(1\right)+7v	\displaystyle =	\displaystyle 168	Substitute b=1
\displaystyle 42+7v	\displaystyle =	\displaystyle 168	Evaluate the product
\displaystyle 7v	\displaystyle =	\displaystyle 126	Subtraction property of equality
\displaystyle v	\displaystyle =	\displaystyle 18	Division property of equality

Idea summary

The standard form of a line is:

\displaystyle Ax+By=C

\bm{A,\,B,\,C}

are integers

To write the equation of the line in standard form, we will follow these steps:

Use the given information to write the equation in slope-intercept form, y=mx+b.
If m is a fraction, multiply each term in the equation by the denominator of m.
Move the x term to the y side of the equation.

Standard form is useful when we know, or want to know both intercepts of the line.

Horizontal and vertical lines

Recall that lines can also be either horizontal or vertical. These types of lines will not look like slope-intercept form. However, they do follow another type of special pattern.

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\,\\\,Horizontal lines are the set of all points with a fixed y-value. They are parallel to the x-axis and have equations of the form y=a, where a is a real number. Horizontal lines have a slope of zero.

Shown is the horizontal line y=2.

Notice this equation is in standard form with {A=0}, since the equation is equivalent to {0x + y = 2}.

What happens to the equation y=mx+b if you substitute 0 for m? We get y=b.

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\,\\\,Vertical lines are the set of all points with a fixed x-value. They are parallel to the y-axis and have equations of the form x=a, where a is a real number. Vertical lines have a slope that is undefined.

Shown is the vertical line x=1.

Notice this equation is in standard form with B=0.

Examples

Example 5

Consider the line x=-8.

Plot the line on a coordinate plane.

Worked Solution

Create a strategy

We can use the fact that vertical lines have equations of the form x=a.

Apply the idea

This will be a vertical line that crosses the x-axis at -8.

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Determine the domain and range of the line.

Worked Solution

Create a strategy

The domain of a vertical line is restricted to the single value of x at which the line is located.

The range of a vertical line is all possible y-values since the line extends infinitely in both the positive and negative y-directions.

Apply the idea

Domain: \{-8\}

Range: (-\infty,\,\infty)

Example 6

Write the equation of the line given.

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Worked Solution

Apply the idea

Horizontal lines have a fixed y-value and have equations of the form y=a.

The fixed y-value for this line is -3, so the equation of the line is

y = -3

An undefined slope through the point (2,\,-1).

Worked Solution

Apply the idea

Lines that have an undefined slope are vertical lines which have equations with the form x=a.

The x-value of the ordered pair (2,\,-1) is 2, so the equation can be written as

x = 2

Idea summary

Horizontal lines are the set of all points with a fixed y-value. They are parallel to the x-axis and have equations of the form y=a, where a is a real number.

Vertical lines are the set of all points with a fixed x-value. They are parallel to the y-axis and have equations of the form x=a, where a is a real number.

Outcomes

A.F.1

The student will investigate, analyze, and compare linear functions algebraically and graphically, and model linear relationships.

A.F.1a

Determine and identify the domain, range, zeros, slope, and intercepts of a linear function, presented algebraically or graphically, including the interpretation of these characteristics in contextual situations.

A.F.1c

Write equivalent algebraic forms of linear functions, including slope-intercept form, standard form, and point-slope form, and analyze and interpret the information revealed by each form.

A.F.1di

Write the equation of a linear function to model a linear relationship between two quantities, including those that can represent contextual situations. Writing the equation of a linear function will include the following situations: i) given the graph of a line;

A.F.1dii

A.F.1diii

A.F.1div

A.F.1dv

A.F.1f

Graph a linear function in two variables, with and without the use of technology, including those that can represent contextual situations.

A.F.1h

Compare and contrast the characteristics of linear functions represented algebraically, graphically, in tables, and in contextual situations.

3.04 Standard form

Ideas

Standard form

Examples

Example 1

Example 2

Example 3

Example 4

Horizontal and vertical lines

Examples

Example 5

Example 6

Outcomes

A.F.1

A.F.1a

A.F.1c

A.F.1di

A.F.1dii

A.F.1diii

A.F.1div

A.F.1dv

A.F.1f

A.F.1h

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