ORCCA Standard Form

Section 4.7 Standard Form

We've seen that a linear relationship can be expressed with an equation in slope-intercept form or with an equation in point-slope-form. There is a third standard form that you can use to write line equations. It's so “standard” that it's actually known as standard form.

Subsection 4.7.1 Standard Form Definition

Imagine trying to gather donations to pay for a $\$10,000$ medical procedure you cannot afford. Oversimplifying the mathematics a bit, suppose that there were only two types of donors in the world: those who will donate $\$20$ and those who will donate $\$100\text{.}$ How many of each, or what combination, do you need to reach the funding goal? As in, if $x$ people donate $\$20$ and $y$ people donate $\$100\text{,}$ what numbers could $x$ and $y$ be? The donors of the first type have collectively donated $20x$ dollars and the donors of the second type have collectively donated $100y$ dollars. So altogether you'd need

\begin{equation*} 20x+100y=10000\text{.} \end{equation*}

This is an example of a line equation in standard form.

Definition 4.7.1. Standard Form.

It is always possible to write an equation for a line in the form

\begin{equation} Ax+By=C\label{yab}\tag{4.7.1} \end{equation}

where $A\text{,}$ $B\text{,}$ and $C$ are three numbers (each of which might be $0\text{,}$ although at least one of $A$ and $B$ must be nonzero). This form of a line equation is called standard form. In the context of an application, the meaning of $A\text{,}$ $B\text{,}$ and $C$ depends on that context. This equation is called standard form perhaps because any line can be written this way, even vertical lines which cannot be written using the two previous forms we've studied.

Checkpoint 4.7.2.

Returning to the example with donations for the medical procedure, let's examine the equation

\begin{equation*} 20x+100y=10000\text{.} \end{equation*}

What units are attached to all of the parts of this equation? Both $x$ and $y$ are numbers of people. The $10000$ is in dollars. Both the $20$ and the $100$ are in dollars per person. Note how both sides of the equation are in dollars. On the right, that fact is clear. On the left, $20x$ is in dollars since $20$ is in dollars per person, and $x$ is in people. The same is true for $100y\text{,}$ and the two dollar amounts $20x$ and $100y$ add to a dollar amount.

What is the slope of the linear relationship? It's not immediately visible since $m$ is not part of the standard form equation. But we can use algebra to isolate $y\text{:}$

\begin{align*} 20x+100y\amp=10000\\ 100y\amp=\highlight{-20x}+10000\\ y\amp=\divideunder{-20x+10000}{100}\\ y\amp=\frac{-20x}{100}+\frac{10000}{100}\\ y\amp=-\frac{1}{5}x+100\text{.} \end{align*}

And we see that the slope is $-\frac{1}{5}\text{.}$ OK, what units are on that slope? As always, the units on slope are $\frac{y\text{-unit}}{x\text{-unit}}\text{.}$ In this case that's $\frac{\text{person}}{\text{person}}\text{,}$ which sounds a little weird and seems like it should be simplified away to unitless. But this slope of $-\frac{1}{5}\frac{\text{person}}{\text{person}}$ is saying that for every one extra person who donates $\$20\text{,}$ you only need $\frac{1}{5}$ fewer people donating $\$100$ to still reach your goal.

What is the $y$-intercept? Since we've already converted the equation into slope-intercept form, we can see that it is at $(0,100)\text{.}$ This tells us that if $0$ people donate $\$20\text{,}$ then you will need $100$ people to each donate $\$100\text{.}$

What does a graph for this line look like? We've already converted into slope-intercept form and we could use that to make the graph. But when given a line in standard form, there is another approach that is often used. Returning to

\begin{equation*} 20x+100y=10000\text{,} \end{equation*}

let's calculate the $y$-intercept and the $x$-intercept. Recall that these are points where the line crosses the $y$-axis and $x$-axis. To be on the $y$-axis means that $x=0\text{,}$ and to be on the $x$-axis means that $y=0\text{.}$ All these zeros make the resulting algebra easy to solve:

\begin{align*} 20x+100y\amp=10000\amp20x+100y\amp=10000\\ 20(\substitute{0})+100y\amp=10000\amp20x+100(\substitute{0})\amp=10000\\ 100y\amp=10000\amp20x\amp=10000\\ y\amp=\divideunder{10000}{100}\amp x\amp=\divideunder{10000}{20}\\ y\amp=100\amp x\amp=500 \end{align*}

So we have a $y$-intercept at $(0,100)$ and an $x$-intercept at $(500,0)\text{.}$ If we plot these, we get to mark especially relevant points given the context, and then drawing a straight line between them gives us Figure 4.7.3.

A Cartesian plot where the x-axis represents $20 donors and the y-axis represents $100 donors; the line has a y-intercept of 100 $100 donors and an x-intercept of 500 $20 donors — Figure 4.7.3.

Subsection 4.7.2 The $x$- and $y$-Intercepts

With a linear relationship (and other types of equations), we are often interested in the $x$-intercept and $y$-intercept because they are important in the context. For example, in Figure 4.7.3, the $x$-intercept implies that if no one donates $\$100\text{,}$ you need $500$ people to donate $\$20$ to get us to $\$10{,}000\text{.}$ And the $y$-intercept implies if no one donates $\$20\text{,}$ you need $100$ people to donate $\$100\text{.}$ Let's look at another example.

Example 4.7.4.

James owns a restaurant that uses about $32$ lb of flour every day. He just purchased 1200 lb of flour. Model the amount of flour that remains $x$ days later with a linear equation, and interpret the meaning of its $x$-intercept and $y$-intercept.

Since the rate of change is constant (-32 lb every day), and we know the initial value, we can model the amount of flour at this restaurant with a slope-intercept equation:

\begin{equation*} y=-32x+1200 \end{equation*}

where $x$ represents the number of days passed since the initial purchase, and $y$ represents the amount of flour left (in lb).

A line's $x$-intercept is in the form of $(x,0)\text{,}$ since to be on the $x$-axis, the $y$-coordinate must be $0\text{.}$ To find this line's $x$-intercept, we substitute $y$ in the equation with $0\text{,}$ and solve for $x\text{:}$

\begin{align*} y\amp=-32x+1200\\ \substitute{0}\amp=-32x+1200\\ 0\subtractright{1200}\amp=-32x\\ -1200\amp=-32x\\ \divideunder{-1200}{-32}\amp=x\\ 37.5\amp=x \end{align*}

So the line's $x$-intercept is at $(37.5,0)\text{.}$ In context this means the flour would last for $37.5$ days.

A line's $y$-intercept is in the form of $(0,y)\text{.}$ This line equation is already in slope-intercept form, so we can just see that its $y$-intercept is at $(0,1200)\text{.}$ In general though, we would substitute $x$ in the equation with $0\text{,}$ and we have:

\begin{align*} y\amp=-32x+1200\\ y\amp=-32(\substitute{0})+1200\\ y\amp=1200 \end{align*}

So yes, the line's $y$-intercept is at $(0,1200)\text{.}$ This means that when the flour was purchased, there was 1200 lb of it. In other words, the $y$-intercept tells us one of the original pieces of information: in the beginning, James purchased 1200 lb of flour.

If a line is in standard form, it's often easiest to graph it using its two intercepts.

Example 4.7.5.

Graph $2x-3y=-6$ using its intercepts. Then use the intercepts to calculate the line's slope.

Explanation

To graph a line by its $x$-intercept and $y$-intercept, it might help to first set up a table like Table 4.7.6:

Table 4.7.6. Intercepts of $2x-3y=-6$

	$x$-value	$y$-value	Points
$x$-intercept		$0$
$y$-intercept	$0$

A table like this might help you stay focused on the fact that we are searching for two points. As we've noted earlier, an $x$-intercept is on the $x$-axis, and so its $y$-coordinate must be $0\text{.}$ This is worth taking special note of: to find an $x$-intercept, $y$ must be $0\text{.}$ This is why we put $0$ in the $y$-value cell of the $x$-intercept. Similarly, a line's $y$-intercept has $x=0\text{,}$ and we put $0$ into the $x$-value cell of the $y$-intercept.

Next, we calculate the line's $x$-intercept by substituting $y=0$ into the equation

\begin{align*} 2x-3y\amp=-6\\ 2x-3(\substitute{0})\amp=-6\\ 2x\amp=-6\\ x\amp=-3 \end{align*}

So the line's $x$-intercept is $(-3,0)\text{.}$

Similarly, we substitute $x=0$ into the equation to calculate the $y$-intercept:

\begin{align*} 2x-3y\amp=-6\\ 2(\substitute{0})-3y\amp=-6\\ -3y\amp=-6\\ y\amp=2 \end{align*}

So the line's $y$-intercept is $(0,2)\text{.}$

Now, while it's true that you only need two points to graph a line, it's a good idea to find a third point as a way to check your work. We call this third point a checkpoint. To find a checkpoint, simply choose an $x$ value, preferably between the two intercepts, or at least not so far from the intercepts that the point would cause you to make a much bigger coordinate system. Plug that $x$ value into the equation for $x$ and then solve for $y\text{.}$ This will give you both the $x$-coordinate and $y$-coordinate of your checkpoint, which you can then plot. If it lines up with the intercepts, then you can be pretty sure that your line is graphed correctly.

Let's find a checkpoint for this line by choosing $x=-2\text{.}$ Notice that this $x$ value falls between the $x$-coordinates of the two intercepts we've already found. Next, we plug that value in for $x$ and solve for $y\text{:}$

\begin{align*} 2x-3y\amp=-6\\ 2(\substitute{-2})-3y\amp=-6\\ -4-3y\amp=-6\\ -3y\amp=-2\\ y\amp=\frac{2}{3} \end{align*}

So the checkpoint is $\left(-2,\frac{2}{3}\right)\text{.}$

Now we can complete the table:

Table 4.7.7. Intercepts and Checkpoint of $2x-3y=-6$

	$x$-value	$y$-value	Points
$x$-intercept	$-3$	$0$	$(-3,0)$
$y$-intercept	$0$	$2$	$(0,2)$
checkpoint	$-2$	$\frac{2}{3}$	$(-2,\frac{2}{3})$

With both the intercepts' and the checkpoint's coordinates, we can graph the line:

a coordinate grid with the graph of line 2x-3y=-6; the x-intercept is (-3,0) and the y-intercept is (0,2) — Figure 4.7.8. Graph of $2x-3y=-6$

There is a slope triangle from the $x$-intercept to the origin up to the $y$-intercept. It tells us that the slope is

\begin{equation*} m=\frac{\Delta y}{\Delta x}=\frac{2}{3}\text{.} \end{equation*}

This last example generalizes to a fact worth noting.

Fact 4.7.9.

If a line's $x$-intercept is at $(r,0)$ and its $y$-intercept is at $(0,b)\text{,}$ then the slope of the line is $-\frac{b}{r}\text{.}$ (Unless the line passes through the origin. Then both $r$ and $b$ equal $0\text{,}$ and then this fraction is undefined and the slope of the line could be anything.)

Checkpoint 4.7.10.

Subsection 4.7.3 Transforming between Standard Form and Slope-Intercept Form

Sometimes a linear equation arises in standard form, but it would be useful to see that equation in slope-intercept form. Or perhaps, vice versa is true.

A linear equation in slope-intercept form tells us important information about the line: its slope $m$ and $y$-intercept $(0,b)\text{.}$ However, a line's standard form does not show those two important values. As a result, we often need to change a line's equation from standard form to slope-intercept form. Let's look at some examples.

Example 4.7.11.

Change $2x-3y=-6$ to slope-intercept form, and then graph it.

Explanation

Since a line in slope-intercept form looks like $y=\ldots\text{,}$ we will solve for $y$ in $2x-3y=-6\text{:}$

\begin{align*} 2x-3y\amp=-6\\ -3y\amp=-6\subtractright{2x}\\ -3y\amp=-2x-6\\ y\amp=\divideunder{-2x-6}{-3}\\ y\amp=\frac{-2x}{-3}-\frac{6}{-3}\\ y\amp=\frac{2}{3}x+2 \end{align*}

In the third line, we wrote $-2x-6$ on the right side, instead of $-6-2x\text{.}$ The only reason we did this is because we are headed to slope-intercept form, where the $x$-term is traditionally written first.

Now we can see that the slope is $\frac{2}{3}$ and the $y$-intercept is at $(0,2)\text{.}$ With these things found, we can graph the line using slope triangles.

Compare this graphing method with the Graphing by Intercepts method in Example 4.7.5. We have more points in this graph, thus we can graph the line more accurately.

a coordinate gris with the graph of the line 2x-3y=-6; the y-intercept is 2 and the slope is 2/3; slope triangles are shown along the line with a run of 3 and a rise of 2 — Figure 4.7.12. Graphing $2x-3y=-6$ with Slope Triangles

Example 4.7.13.

Graph $2x-3y=0\text{.}$

Explanation

First, we will try (and fail) to graph this line using its $x$- and $y$-intercepts.

Trying to find the $x$-intercept:

\begin{align*} 2x-3y\amp=0\\ 2x-3(\substitute{0})\amp=0\\ 2x\amp=0\\ x\amp=0 \end{align*}

So the line's $x$-intercept is at $(0,0)\text{,}$ the origin.

Huh, that is also on the $y$-axis…

Trying to find the $y$-intercept:

\begin{align*} 2x-3y\amp=0\\ 2(\substitute{0})-3y\amp=0\\ -3y\amp=0\\ y\amp=0 \end{align*}

So the line's $y$-intercept is also at $(0,0)\text{.}$

Since both intercepts are the same point, there is no way to use the intercepts alone to graph this line. So what can be done?

Several approaches are out there, but one is to convert the line equation into slope-intercept form:

\begin{align*} 2x-3y\amp=0\\ -3y\amp=0\subtractright{2x}\\ -3y\amp=-2x\\ y\amp=\divideunder{-2x}{-3}\\ y\amp=\frac{2}{3}x \end{align*}

So the line's slope is $\frac{2}{3}\text{,}$ and we can graph the line using slope triangles and the intercept at $(0,0)\text{,}$ as in Figure 4.7.14.

a coordinate grid with the graph of line 2x-3y=0; The line has a y-intercept of 0 and a slope of 2/3; slope triangels are shown along the line with a run of 3 and a rise of 2 — Figure 4.7.14. Graphing $2x-3y=0$ with Slope Triangles

In summary, if $C=0$ in a standard form equation, it's convenient to graph it by first converting the equation to slope-intercept form.

Example 4.7.15.

Write the equation $y=\frac{2}{3}x+2$ in standard form.

Explanation

Once we subtract $\frac{2}{3}x$ on both sides of the equation, we have

\begin{equation*} -\frac{2}{3}x+y=2 \end{equation*}

Technically, this equation is already in standard form $Ax+By=C\text{.}$ However, you might like to end up with an equation that has no fractions, and so you can take some extra steps.

\begin{align*} y\amp=\frac{2}{3}x+2\\ y\subtractright{\frac{2}{3}x}\amp=2\\ -\frac{2}{3}x+y\amp=2\\ \end{align*}

This is in standard form, but we keep going to clear away the fraction.

\begin{align*} \multiplyleft{3}\left(-\frac{2}{3}x+y\right)\amp=\multiplyleft{3}2\\ -2x+3y\amp=6 \end{align*}