# 5.2: Forms of Linear Equations

**At Grade**Created by: CK-12

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**Practice**Forms of Linear Equations

In this Concept, you will learn how to write equations in standard form.

### Watch This

CK-12 Foundation: 0502S Standard Form of Linear Equations (H264)

### Try This

Now that you’ve worked with equations in all three basic forms, check out the Java applet at http://www.ronblond.com/M10/lineAP/index.html. You can use it to manipulate graphs of equations in all three forms, and see how the graphs change when you vary the terms of the equations.

### Guidance

You’ve already encountered another useful form for writing linear equations: **standard form.** An equation in standard form is written \begin{align*}ax+by=c\end{align*}, where \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} are all integers and \begin{align*}a\end{align*} is positive. (Note that the \begin{align*}b\end{align*} in the standard form is different than the \begin{align*}b\end{align*} in the slope-intercept form.)

One useful thing about standard form is that it allows us to write equations for vertical lines, which we can’t do in slope-intercept form.

For example, let’s look at the line that passes through points (2, 6) and (2, 9). How would we find an equation for that line in slope-intercept form?

First we’d need to find the slope: \begin{align*}m=\frac{9-6}{0-0}=\frac{3}{0}\end{align*}. But that slope is undefined because we can’t divide by zero. And if we can’t find the slope, we can’t use point-slope form either.

If we just graph the line, we can see that \begin{align*}x\end{align*} equals 2 no matter what \begin{align*}y\end{align*} is. There’s no way to express that in slope-intercept or point-slope form, but in standard form we can just say that \begin{align*}x+0y=2\end{align*}, or simply \begin{align*}x=2\end{align*}.

**Converting to Standard Form**

To convert an equation from another form to standard form, all you need to do is rewrite the equation so that all the variables are on one side of the equation and the coefficient of \begin{align*}x\end{align*} is not negative.

#### Example A

*Rewrite the following equations in standard form:*

a) \begin{align*}y=5x-7\end{align*}

b) \begin{align*}y-2=-3(x+3)\end{align*}

c) \begin{align*}y=\frac{2}{3}x+\frac{1}{2}\end{align*}

**Solution**

We need to rewrite each equation so that all the variables are on one side and the coefficient of \begin{align*}x\end{align*} is not negative.

a) \begin{align*}y=5x-7\end{align*}

Subtract \begin{align*}y\end{align*} from both sides to get \begin{align*}0=5x-y-7\end{align*}.

Add 7 to both sides to get \begin{align*}7=5x-y\end{align*}.

Flip the equation around to put it in standard form: \begin{align*}5x-y=7\end{align*}.

b) \begin{align*}y-2=-3(x+3)\end{align*}

Distribute the –3 on the right-hand-side to get \begin{align*}y-2=-3x-9\end{align*}.

Add \begin{align*}3x\end{align*} to both sides to get \begin{align*}y+3x-2=-9\end{align*}.

Add 2 to both sides to get \begin{align*}y+3x=-7\end{align*}. Flip that around to get \begin{align*}3x+y=-7\end{align*}.

c) \begin{align*}y=\frac{2}{3}x+\frac{1}{2}\end{align*}

Find the common denominator for all terms in the equation – in this case that would be 6.

Multiply all terms in the equation by 6: \begin{align*}6 \left(y=\frac{2}{3}x+\frac{1}{2}\right) \Rightarrow 6y=4x+3\end{align*}

Subtract \begin{align*}6y\end{align*} from both sides: \begin{align*}0=4x-6y+3\end{align*}

Subtract 3 from both sides: \begin{align*}-3=4x-6y\end{align*}

The equation in standard form is \begin{align*}4x-6y=-3\end{align*}.

**Graphing Equations in Standard Form**

When an equation is in slope-intercept form or point-slope form, you can tell right away what the slope is. How do you find the slope when an equation is in standard form?

Well, you could rewrite the equation in slope-intercept form and read off the slope. But there’s an even easier way. Let’s look at what happens when we rewrite an equation in standard form.

Starting with the equation \begin{align*}ax+by=c\end{align*}, we would subtract \begin{align*}ax\end{align*} from both sides to get \begin{align*}by=-ax+c\end{align*}. Then we would divide all terms by \begin{align*}b\end{align*} and end up with \begin{align*}y=-\frac{a}{b}x+\frac{c}{b}\end{align*}.

That means that the slope is \begin{align*}-\frac{a}{b}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}\end{align*}. So next time we look at an equation in standard form, we don’t have to rewrite it to find the slope; we know the slope is just \begin{align*}-\frac{a}{b}\end{align*}, where \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the coefficients of \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the equation.

#### Example B

*Find the slope and the \begin{align*}y-\end{align*}intercept of the following equations written in standard form.*

a) \begin{align*}3x+5y=6\end{align*}

b) \begin{align*}2x-3y=-8\end{align*}

c) \begin{align*}x-5y=10\end{align*}

**Solution**

a) \begin{align*}a=3, b=5\end{align*}, and \begin{align*}c=6\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=-\frac{3}{5}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{6}{5}\end{align*}.

b) \begin{align*}a=2, b=-3\end{align*}, and \begin{align*}c=-8\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=\frac{2}{3}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{8}{3}\end{align*}.

c) \begin{align*}a=1, b=-5\end{align*}, and \begin{align*}c=10\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=\frac{1}{5}\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{10}{-5}=-2\end{align*}.

Once we’ve found the slope and \begin{align*}y-\end{align*}intercept of an equation in standard form, we can graph it easily. But if we start with a graph, how do we find an equation of that line in standard form?

First, remember that we can also use the cover-up method to graph an equation in standard form, by finding the intercepts of the line. For example, let’s graph the line given by the equation \begin{align*}3x-2y=6\end{align*}.

To find the \begin{align*}x-\end{align*}intercept, cover up the \begin{align*}y\end{align*} term (remember, the \begin{align*}x-\end{align*}intercept is where \begin{align*}y = 0\end{align*}):

\begin{align*}3x=6 \Rightarrow x=2\end{align*}

The \begin{align*}x-\end{align*} intercept is (2, 0).

To find the \begin{align*}y-\end{align*}intercept, cover up the \begin{align*}x\end{align*} term (remember, the \begin{align*}y-\end{align*}intercept is where \begin{align*}x = 0)\end{align*}:

\begin{align*}-2y=6 \Rightarrow y=-3\end{align*}

The \begin{align*}y-\end{align*}intercept is (0, -3).

We plot the intercepts and draw a line through them that extends in both directions:

Now we want to apply this process in reverse—to start with the graph of the line and write the equation of the line in standard form.

#### Example C

*Find the equation of each line and write it in standard form.*

a)

b)

c)

**Solution**

a) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}(3, 0) \Rightarrow x=3\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, -4) \Rightarrow y=-4\end{align*}

We saw that in standard form \begin{align*}ax+by=c\end{align*}: if we “cover up” the \begin{align*}y\end{align*} term, we get \begin{align*}ax = c\end{align*}, and if we “cover up” the \begin{align*}x\end{align*} term, we get \begin{align*}by = c\end{align*}.

So we need to find values for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} so that we can plug in 3 for \begin{align*}x\end{align*} and -4 for \begin{align*}y\end{align*} and get the same value for \begin{align*}c\end{align*} in both cases. This is like finding the least common multiple of the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}intercepts.

In this case, we see that multiplying \begin{align*}x=3\end{align*} by 4 and multiplying \begin{align*}y=-4\end{align*} by –3 gives the same result:

\begin{align*}(x=3) \times 4 \Rightarrow 4x=12 \quad \text{and} \quad (y=-4) \times (-3) \Rightarrow -3y=12\end{align*}

Therefore, \begin{align*}a = 4, b = -3\end{align*} and \begin{align*}c = 12\end{align*} and **the equation in standard form is** \begin{align*}4x-3y=12\end{align*}.

b) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}(3, 0) \Rightarrow x=3\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, 3) \Rightarrow y=3\end{align*}

The values of the intercept equations are already the same, so \begin{align*}a = 1, b = 1\end{align*} and \begin{align*}c = 3\end{align*}. **The equation in standard form is** \begin{align*}x+y=3\end{align*}.

c) We see that the \begin{align*}x-\end{align*}intercept is \begin{align*}\left(\frac{3}{2}, 0 \right) \Rightarrow x=\frac{3}{2}\end{align*} and the \begin{align*}y-\end{align*}intercept is \begin{align*}(0, 4) \Rightarrow y=4\end{align*}

Let’s multiply the \begin{align*}x-\end{align*}intercept equation by \begin{align*}2 \Rightarrow 2x=3\end{align*}

Then we see we can multiply the \begin{align*}x-\end{align*}intercept again by 4 and the \begin{align*}y-\end{align*}intercept by 3, so we end up with \begin{align*}8x=12\end{align*} and \begin{align*}3y=12\end{align*}.

**The equation in standard form is** \begin{align*}8x+3y=12\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Standard Form of Linear Equations

### Vocabulary

- An equation in
**standard form**is written \begin{align*}ax+by=c\end{align*}, where \begin{align*}a, b\end{align*}, and \begin{align*}c\end{align*} are all integers and \begin{align*}a\end{align*} is positive. (Note that the \begin{align*}b\end{align*} in the standard form is different than the \begin{align*}b\end{align*} in the slope-intercept form.)

### Guided Practice

a) \begin{align*}10x+2y=5\end{align*}

b) \begin{align*}21x-3y=-9\end{align*}

**Solution:**

a) \begin{align*}a=10, b=2\end{align*}, and \begin{align*}c=5\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=-\frac{10}{2}=-5\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{5}{2}=2.5\end{align*}.

b) \begin{align*}a=21, b=-3\end{align*}, and \begin{align*}c=-9\end{align*}, so the slope is \begin{align*}-\frac{a}{b}=-\frac{21}{-3}=7\end{align*}, and the \begin{align*}y-\end{align*}intercept is \begin{align*}\frac{c}{b}=\frac{-9}{-3}=3\end{align*}.

### Practice

For 1-6, rewrite the following equations in standard form.

- \begin{align*}y=3x-8\end{align*}
- \begin{align*}y-7=-5(x-12)\end{align*}
- \begin{align*}2y=6x+9\end{align*}
- \begin{align*}y=\frac{9}{4}x+\frac{1}{4}\end{align*}
- \begin{align*}y+\frac{3}{5}=\frac{2}{3}(x-2)\end{align*}
- \begin{align*}3y+5=4(x-9)\end{align*}

For 7-12, find the slope and \begin{align*}y-\end{align*}intercept of the following lines.

- \begin{align*}5x-2y=15\end{align*}
- \begin{align*}3x+6y=25\end{align*}
- \begin{align*}x-8y=12\end{align*}
- \begin{align*}3x-7y=20\end{align*}
- \begin{align*}9x-9y=4\end{align*}
- \begin{align*}6x+y=3\end{align*}

For 13-14, find the equation of each line and write it in standard form.

### Image Attributions

## Description

## Learning Objectives

Here you'll learn how to write equations of lines in the standard form of @$\begin{align*}ax + by = c\end{align*}@$. You'll also learn how to find the slope and @$\begin{align*}y\end{align*}@$-intercept of lines written in standard form.