# 11.2 Graphing linear equations  (Page 3/6)

 Page 3 / 6

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 $\left(0,b\right).$

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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