$\require{cancel}\newcommand\degree[0]{^{\circ}} \newcommand\Ccancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} \newcommand{\alert}[1]{\boldsymbol{\color{magenta}{#1}}} \newcommand{\blert}[1]{\boldsymbol{\color{blue}{#1}}} \newcommand{\bluetext}[1]{\color{blue}{#1}} \delimitershortfall-1sp \newcommand\abs[1]{\left|#1\right|} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$

## SectionGraphs of Linear Equations

At the beginning of this chapter, we defined linear equations to be equations in which each term was either a constant or a constant times a single variable. In this section, we will look at linear equations in two variables, and, using the skills from the previous section, explores their graphs.

###### In this section, you will...
• generate tables to represent linear equations in two variables

• graph linear equations from a table

• graph lines from their equations

### SubsectionGraphing Linear Equations

Consider the linear equation $8x+4y=12\text{.}$ Before we try to graph this equation, let's generate a table of ordered pairs for this relation. To do so, we can pick any $x$ values and find their corresponding $y$ values, or vice versa. Some ordered pairs are shown below.

 $x$ $y$ Ordered pair $-2$ $7$ $(-2,7)$ $-1$ $5$ $(-1,5)$ $0$ $3$ $(0,3)$ $4$ $-5$ $(4,-5)$ $6$ $-9$ $(6,-9)$

We can now plot these points. All of these points fall in a straight line; connecting the points, we get the graph of the equation $8x+4y=12\text{.}$ We add arrows on either end to indicate that the graph extends indefinitely in both directions.

Let's create a table and graph the linear equation $2y-x=8\text{.}$ Again, we start by generating a table with several ordered pairs that satisfy this equation.

 $x$ $y$ Ordered pair $-3$ $2.5$ $(-3,2.5)$ $-1$ $3.5$ $(-1,3.5)$ $0$ $4$ $(0,4)$ $3$ $5.5$ $(4,5.5)$ $2$ $5$ $(2,5)$

Notice that several of the ordered pairs in our table have decimal-valued coordinates. If you are randomly selecting values of $x\text{,}$ this will often happen and is okay. However, if you are plotting these points by hand and cannot accurately plot decimal-valued points, you should avoid using these points. In the table above, we found two points with integer coordinates: $(0,4)$ and $(2,5)\text{.}$ These two points are enough to graph this equation. We can then verify that all of the decimal-valued coordinates do in fact fall on this line.

### SubsectionIntercepts of Graphs of Linear Equations

It is often useful to describe where the graph of a linear equation meets the $x$- and $y$-axes. The point where a graph intersects the $x$-axis is called the $x$-intercept of the graph, and the point where a graph intersects the $y$-axis it called the $y$-intercept.

###### Example79

Look back at the graphs of $8x+4y=12$ and $2y-x=8$ from earlier in this section. What are the $y$-intercepts of each of these graphs?

Solution

The $y$-intercept of $8x+4y=12$ is $(0,4)\text{.}$ The $y$-intercept of $2y-x=8$ is $(0,4)\text{.}$

It is not always clear from the graph what the intercepts of a line are. For example, consider the graph of the equation $6y-3x=-17\text{.}$

It is not clear where exactly the line is intersecting the axes. Although we could estimate where these points lie, there is a more accurate way to find the intercepts.

Think about points that lie on the $x$-axis. What do these points have in common? If a point lies on the $x$-axis, it will have the form $(x,0)\text{;}$ the $y$-value will always be $0\text{.}$ This means that to find the $x$-intercept of a graph of a linear equation, we are looking for the $x$-value that corresponds to $y=0\text{.}$ Substituting $y=0$ into $6y-3x=-17\text{,}$ we have

\begin{equation*} \begin{aligned} 6(\alert{0})-3x \amp= -17 \\ -3x \amp= -17 \\ x \amp= \dfrac{17}{3} \end{aligned} \end{equation*}

This means that the $x$-intercept of the graph of $6y-3x=-17$ is $(\frac{17}{3},0)\text{.}$

We can think about the $y$-intercepts similarly. Points on the $y$-axis have the form $(0,y)\text{,}$ so they always have $x=0\text{.}$ To find the $y$-intercept of the graph of $6y-3x=-17\text{,}$ we substitute $x=0\text{:}$

\begin{equation*} \begin{aligned} 6y-3(\alert{0}) \amp= -17 \\ 6y \amp= -17 \\ x \amp= -\dfrac{17}{6} \end{aligned} \end{equation*}

This means that the $y$-intercept of the graph of $6y-3x=-17$ is $(0,-\frac{17}{6})\text{.}$

###### Note80

Some lines don't have an $x$-intercept, and some lines don't have a $y$-intercept. Can you draw such lines? These lines will be discussed in the next section.

###### Example81

Find the $x$- and $y$-intercepts of the graph of $6x+9y=18\text{.}$ Verify your answers by graphing the equation.

Solution

To find the $x$-intercept, we substitute $y=0\text{:}$

\begin{equation*} \begin{aligned} 6x+9(0) \amp= 18 \\ 6x \amp= 18 \\ x \amp= 3 \end{aligned} \end{equation*}

This means that the $x$-intercept is at $(3,0)\text{.}$ For the $y$-intercept, we substitute $x=0\text{:}$

\begin{equation*} \begin{aligned} 6(0)+9y \amp= 18 \\ 9y \amp= 18 \\ y \amp= 2 \end{aligned} \end{equation*}

Therefore the $y$-intercept is at $(0,2)\text{.}$