# 11.2 Graphing linear equations  (Page 3/6)

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Graph the equation: $x+y=-2.$

Graph the equation: $x-y=6.$

In the previous example, the three points we found were easy to graph. But this is not always the case. Let’s see what happens in the equation $2x+y=3.$ If $y$ is $0,$ what is the value of $x?$

The solution is the point $\left(\frac{3}{2},0\right).$ This point has a fraction for the $x$ -coordinate. While we could graph this point, it is hard to be precise graphing fractions. Remember in the example $y=\frac{1}{2}x+3,$ we carefully chose values for $x$ so as not to graph fractions at all. If we solve the equation $2x+y=3$ for $y,$ it will be easier to find three solutions to the equation.

$2x+y=3$
$y=-2x+3$

Now we can choose values for $x$ that will give coordinates that are integers. The solutions for $x=0,x=1,$ and $x=-1$ are shown.

$y=-2x+3$
$x$ $y$ $\left(x,y\right)$
$0$ $5$ $\left(-1,5\right)$
$1$ $3$ $\left(0,3\right)$
$-1$ $1$ $\left(1,1\right)$

Graph the equation $3x+y=-1.$

## Solution

Find three points that are solutions to the equation.

First, solve the equation for $y.$

$\begin{array}{ccc}\hfill 3x+y& =& -1\hfill \\ \hfill y& =& -3x-1\hfill \end{array}$

We’ll let $x$ be $0,1,$ and $-1$ to find three points. The ordered pairs are shown in the table. Plot the points, check that they line up, and draw the line.

$y=-3x-1$
$x$ $y$ $\left(x,y\right)$
$0$ $-1$ $\left(0,-1\right)$
$1$ $-4$ $\left(1,-4\right)$
$-1$ $2$ $\left(-1,2\right)$

If you can choose any three points to graph a line, how will you know if your graph matches the one shown in the answers in the book? If the points where the graphs cross the $x\text{-}$ and $y$ -axes are the same, the graphs match.

Graph each equation: $2x+y=2.$

Graph each equation: $4x+y=-3.$

## Graph vertical and horizontal lines

Can we graph an equation with only one variable? Just $x$ and no $y,$ or just $y$ without an $x?$ How will we make a table of values to get the points to plot?

Let’s consider the equation $x=-3.$ The equation says that $x$ is always equal to $-3,$ so its value does not depend on $y.$ No matter what $y$ is, the value of $x$ is always $-3.$

To make a table of solutions, we write $-3$ for all the $x$ values. Then choose any values for $y.$ Since $x$ does not depend on $y,$ you can chose any numbers you like. But to fit the size of our coordinate graph, we’ll use $1,2,$ and $3$ for the $y$ -coordinates as shown in the table.

$x=-3$
$x$ $y$ $\left(x,y\right)$
$-3$ $1$ $\left(-3,1\right)$
$-3$ $2$ $\left(-3,2\right)$
$-3$ $3$ $\left(-3,3\right)$

Then plot the points and connect them with a straight line. Notice in [link] that the graph is a vertical line    .

## Vertical line

A vertical line is the graph of an equation that can be written in the form $x=a.$

The line passes through the $x$ -axis at $\left(a,0\right)$ .

Graph the equation $x=2.$ What type of line does it form?

## Solution

The equation has only variable, $x,$ and $x$ is always equal to $2.$ We make a table where $x$ is always $2$ and we put in any values for $y.$

$x=2$
$x$ $y$ $\left(x,y\right)$
$2$ $1$ $\left(2,1\right)$
$2$ $2$ $\left(2,2\right)$
$2$ $3$ $\left(2,3\right)$

Plot the points and connect them as shown.

The graph is a vertical line passing through the $x$ -axis at $2.$

Graph the equation: $x=5.$

Graph the equation: $x=-2.$

What if the equation has $y$ but no $x$ ? Let’s graph the equation $y=4.$ This time the $y$ -value is a constant, so in this equation $y$ does not depend on $x.$

To make a table of solutions, write $4$ for all the $y$ values and then choose any values for $x.$

We’ll use $0,2,$ and $4$ for the $x$ -values.

$y=4$
$x$ $y$ $\left(x,y\right)$
$0$ $4$ $\left(0,4\right)$
$2$ $4$ $\left(2,4\right)$
$4$ $4$ $\left(4,4\right)$

Plot the points and connect them, as shown in [link] . This graph is a horizontal line    passing through the $y\text{-axis}$ at $4.$

## Horizontal line

A horizontal line is the graph of an equation that can be written in the form $y=b.$

The line passes through the $y\text{-axis}$ at $\left(0,b\right).$

a perfect square v²+2v+_
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