# 6.2 Graphing linear inequalities  (Page 3/3)

 Page 3 / 3

## Practice set a

Graph $3x+y=3$ using the intercept method.

When $x=0$ , $y=3$ ; when $y=0$ , $x=1$

## Graphing using any two or more points

The graphs we have constructed so far have been done by finding two particular points, the intercepts. Actually, any two points will do. We chose to use the intercepts because they are usually the easiest to work with. In the next example, we will graph two equations using points other than the intercepts. We’ll use three points, the extra point serving as a check.

## Sample set b

$x-3y=-10$ .
We can find three points by choosing three $x\text{-values}$ and computing to find the corresponding $y\text{-values}$ . We’ll put our results in a table for ease of reading.

Since we are going to choose $x\text{-values}$ and then compute to find the corresponding $y\text{-values}$ , it will be to our advantage to solve the given equation for $y$ .

$\begin{array}{rrrr}\hfill x-3y& \hfill =& -10\hfill & \hfill \text{Subtract}\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{from both sides}\text{.}\\ \hfill -3y& \hfill =& \hfill -x-10& \hfill \text{Divide both sides by}-3.\\ \hfill y& \hfill =& \hfill \frac{1}{3}x+\frac{10}{3}& \hfill \end{array}$

 $x$ $y$ $\left(x,\text{\hspace{0.17em}}y\right)$ 1 If $x=1$ , then $y=\frac{1}{3}\left(1\right)+\frac{10}{3}=\frac{1}{3}+\frac{10}{3}=\frac{11}{3}$ $\left(1,\text{\hspace{0.17em}}\frac{11}{3}\right)$ $-3$ If $x=-3$ , then $y=\frac{1}{3}\left(-3\right)+\frac{10}{3}=-1+\frac{10}{3}=\frac{7}{3}$ $\left(-3,\frac{7}{3}\right)$ 3 If $x=3$ , then $y=\frac{1}{3}\left(3\right)+\frac{10}{3}=1+\frac{10}{3}=\frac{13}{3}$ $\left(3,\text{\hspace{0.17em}}\frac{13}{3}\right)$

Thus, we have the three ordered pairs (points), $\left(1,\text{\hspace{0.17em}}\frac{11}{3}\right)$ , $\left(-3,\text{\hspace{0.17em}}\frac{7}{3}\right)$ , $\left(3,\text{\hspace{0.17em}}\frac{13}{3}\right)$ . If we wish, we can change the improper fractions to mixed numbers, $\left(1,3\text{\hspace{0.17em}}\frac{2}{3}\right)$ , $\left(-3,2\text{\hspace{0.17em}}\frac{1}{3}\right)$ , $\left(3,4\text{\hspace{0.17em}}\frac{1}{3}\right)$ .

$4x+4y=0$

We solve for $y$ .

$\begin{array}{ccc}4y& =& -4x\\ y& =& -x\end{array}$

 $x$ $y$ $\left(x,\text{\hspace{0.17em}}y\right)$ 0 0 $\left(0,\text{\hspace{0.17em}}0\right)$ 2 $-2$ $\left(2,\text{\hspace{0.17em}}-2\right)$ $-3$ 3 $\left(-3,\text{\hspace{0.17em}}3\right)$

Notice that the $x-$ and $y\text{-intercepts}$ are the same point. Thus the intercept method does not provide enough information to construct this graph.

When an equation is given in the general form $ax+by=c$ , usually the most efficient approach to constructing the graph is to use the intercept method, when it works.

## Practice set b

Graph the following equations.

$x-5y=5$

$x+2y=6$

$2x+y=1$

## Slanted, horizontal, and vertical lines

In all the graphs we have observed so far, the lines have been slanted. This will always be the case when both variables appear in the equation. If only one variable appears in the equation, then the line will be either vertical or horizontal. To see why, let’s consider a specific case:

Using the general form of a line, $ax+by=c$ , we can produce an equation with exactly one variable by choosing $a=0$ , $b=5$ , and $c=15$ . The equation $ax+by=c$ then becomes

$0x+5y=15$

Since $0\cdot \left(\text{any}\text{\hspace{0.17em}}\text{number}\right)=0$ , the term $0x$ is $0$ for any number that is chosen for $x$ .

Thus,

$0x+5y=15$

becomes

$0+5y=15$

But, $0$ is the additive identity and $0+5y=5y$ .

$5y=15$

Then, solving for $y$ we get

$y=3$

This is an equation in which exactly one variable appears.

This means that regardless of which number we choose for $x$ , the corresponding $y\text{-value}$ is 3. Since the $y\text{-value}$ is always the same as we move from left-to-right through the $x\text{-values}$ , the height of the line above the $x\text{-axis}$ is always the same (in this case, 3 units). This type of line must be horizontal.

An argument similar to the one above will show that if the only variable that appears is $x$ , we can expect to get a vertical line.

## Sample set c

Graph $y=4$ .
The only variable appearing is $y$ . Regardless of which $x\text{-value}$ we choose, the $y\text{-value}$ is always 4. All points with a $y\text{-value}$ of 4 satisfy the equation. Thus we get a horizontal line 4 unit above the $x\text{-axis}$ .

 $x$ $y$ $\left(x,\text{\hspace{0.17em}}y\right)$ $-3$ 4 $\left(-3,\text{\hspace{0.17em}}4\right)$ $-2$ 4 $\left(-2,\text{\hspace{0.17em}}4\right)$ $-1$ 4 $\left(-1,\text{\hspace{0.17em}}4\right)$ 0 4 $\left(0,\text{\hspace{0.17em}}4\right)$ 1 4 $\left(1,\text{\hspace{0.17em}}4\right)$ 2 4 $\left(2,\text{\hspace{0.17em}}4\right)$ 3 4 $\left(3,\text{\hspace{0.17em}}4\right)$ 4 4 $\left(4,\text{\hspace{0.17em}}4\right)$

Graph $x=-2$ .
The only variable that appears is $x$ . Regardless of which $y\text{-value}$ we choose, the $x\text{-value}$ will always be $-2$ . Thus, we get a vertical line two units to the left of the $y\text{-axis}$ .

 $x$ $y$ $\left(x,\text{}y\right)$ $-2$ $-4$ $\left(-2,\text{\hspace{0.17em}}-4\right)$ $-2$ $-3$ $\left(-2,\text{\hspace{0.17em}}-3\right)$ $-2$ $-2$ $\left(-2,\text{\hspace{0.17em}}-2\right)$ $-2$ $-1$ $\left(-2,\text{\hspace{0.17em}}-1\right)$ $-2$ 0 $\left(-2,\text{\hspace{0.17em}}0\right)$ $-2$ 1 $\left(-2,\text{\hspace{0.17em}}1\right)$ $-2$ 2 $\left(-2,\text{\hspace{0.17em}}0\right)$ $-2$ 3 $\left(-2,\text{\hspace{0.17em}}3\right)$ $-2$ 4 $\left(-2,\text{\hspace{0.17em}}4\right)$

## Practice set c

Graph $y=2$ .

Graph $x=-4$ .

## Summarizing our results we can make the following observations:

1. When a linear equation in two variables is written in the form $ax+by=c$ , we say it is written in general form .
2. To graph an equation in general form it is sometimes convenient to use the intercept method.
3. A linear equation in which both variables appear will graph as a slanted line.
4. A linear equation in which only one variable appears will graph as either a vertical or horizontal line.

$x=a$ graphs as a vertical line passing through $a$ on the $x\text{-axis}$ .
$y=b$ graphs as a horizontal line passing through $b$ on the $y\text{-axis}$ .

## Exercises

For the following problems, graph the equations.

$-3x+y=-1$

$3x-2y=6$

$-2x+y=4$

$x-3y=5$

$2x-3y=6$

$2x+5y=10$

$3\left(x-y\right)=9$

$-2x+3y=-12$

$y+x=1$

$4y-x-12=0$

$2x-y+4=0$

$-2x+5y=0$

$y-5x+4=0$

$0x+y=3$

$0x+2y=2$

$0x+\frac{1}{4}y=1$

$4x+0y=16$

$\frac{1}{2}x+0y=-1$

$\frac{2}{3}x+0y=1$

$x=\frac{3}{2}$

$y=3$

$y=-2$

$y=-2$

$-4y=20$

$x=-4$

$-3x=-9$

$-x+4=0$

Construct the graph of all the points that have coordinates $\left(a,\text{\hspace{0.17em}}a\right)$ , that is, for each point, the $x-$ and $y\text{-values}$ are the same.

## Calculator problems

$2.53x+4.77y=8.45$

$1.96x+2.05y=6.55$

$4.1x-6.6y=15.5$

$626.01x-506.73y=2443.50$

## Exercises for review

( [link] ) Name the property of real numbers that makes $4+x=x+4$ a true statement.

( [link] ) Supply the missing word. The absolute value of a number $a$ , denoted $|a|$ , is the from $a$ to $0$ on the number line.

( [link] ) Find the product $\left(3x+2\right)\left(x-7\right)$ .

$3{x}^{2}-19x-14$

( [link] ) Solve the equation $3\left[3\left(x-2\right)+4x\right]-24=0$ .

( [link] ) Supply the missing word. The coordinate axes divide the plane into four equal regions called .

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