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We start by finding three points that are solutions to the equation. We can choose any value for x or y , and then solve for the other variable.

Since y is isolated on the left side of the equation, it is easier to choose values for x . We will use 0 , 1 , and -2 for x for this example. We substitute each value of x into the equation and solve for y .

The figure shows three algebraic substitutions into an equation. The first substitution is for x = -2, with -2 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses -2, shown in blue, closed parentheses, + 1. The next line is y = - 4 + 1. The next line is y = -3. The last line is “ordered pair -2, -3”. The second  substitution is for x = 0, with 0 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses 0, shown in blue, closed parentheses, + 1. The next line is y = 0 + 1. The next line is y = 1. The last line is “ordered pair 0, 2”. The third substitution is for x = 1, with 1 shown in blue. The next line is y = 2 x + 1. The next line is y = 2 open parentheses 1, shown in blue, closed parentheses, + 1. The next line is y = 2 + 1. The next line is y = 3. The last line is “ordered pair -1, 3”.

We can organize the solutions in a table. See [link] .

y = 2 x + 1
x y ( x , y )
0 1 ( 0 , 1 )
1 3 ( 1 , 3 )
−2 −3 ( −2 , −3 )

Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work. See [link] .

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. Three labeled points are shown, “ordered pair -2, -3”, “ordered pair 0, 1”, and ordered pair 1, 3”.

Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of y = 2 x + 1 .

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -2, -3”, “ordered pair 0, 1”, and ordered pair 1, 3”.

Graph a linear equation by plotting points.

  1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
  2. Plot the points on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
  3. Draw the line through the points. Extend the line to fill the grid and put arrows on both ends of the line.

It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you plot only two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line. If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. See [link] .

There are two figures. Figure a shows three points that are all contained on a straight line. There is a line with arrows that passed through the three points. Figure b shows 3 points that are not all arranged in a straight line.
Look at the difference between (a) and (b). All three points in (a) line up so we can draw one line through them. The three points in (b) do not line up. We cannot draw a single straight line through all three points.

Graph the equation y = −3 x .

Solution

Find three points that are solutions to the equation. It’s easier to choose values for x , and solve for y . Do you see why?

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses 0, shown in blue, closed parentheses. The next line is y = 0. The last line is “ordered pair 0, 0 “. The second substitution is for x = 1, with 0 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses 1, shown in blue, closed parentheses. The next line is y = -3. The last line is “ordered pair 1, -3”. The third substitution is for x = -2, with -2 shown in blue. The next line is y = -3 x. The next line is y = -3 open parentheses -2, shown in blue, closed parentheses. The next line is y = 6. The last line is “ordered pair -2, 6 “.

List the points in a table.

y = −3 x
x y ( x , y )
0 0 ( 0 , 0 )
1 3 ( 1 , −3 )
−2 6 ( −2 , 6 )

Plot the points, check that they line up, and draw the line as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair -2, 6”, “ordered pair 0, 0”, and ordered pair 1, -3”. The line is labeled y = -3 x.
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Graph the equation by plotting points: y = −4 x .


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 0” and “ordered pair 4, -4”.

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Graph the equation by plotting points: y = x .


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 0” and “ordered pair 1, -4”.

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When an equation includes a fraction as the coefficient of x , we can substitute any numbers for x . But the math is easier if we make ‘good’ choices for the values of x . This way we will avoid fraction answers, which are hard to graph precisely.

Graph the equation y = 1 2 x + 3 .

Solution

Find three points that are solutions to the equation. Since this equation has the fraction 1 2 as a coefficient of x , we will choose values of x carefully. We will use zero as one choice and multiples of 2 for the other choices.

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 0, shown in blue, closed parentheses, + 3.  The next line is y = 3. The last line is “ordered pair 0, 3”. The second substitution is for x = 2, with 2 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 2, shown in blue, closed parentheses, + 3.  The next line is y = 4. The last line is “ordered pair 2, 4”. The third substitution is for x = 4, with 4 shown in blue. The next line is y = 1 over 2 x + 3. The next line is y = 1 over 2 open parentheses 4, shown in blue, closed parentheses, + 3.  The next line is y = 5. The last line is “ordered pair 4, 5”.

The points are shown in the table.

y = 1 2 x + 3
x y ( x , y )
0 3 ( 0 , 3 )
2 4 ( 2 , 4 )
4 5 ( 4 , 5 )

Plot the points, check that they line up, and draw the line as shown.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair 0, 3”, “ordered pair 2, 4”, and ordered pair 4, 5”. The line is labeled y = 1 over 2 x + 3.
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Graph the equation: y = 1 3 x 1 .


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, -1” and “ordered pair 3, 0”.

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Graph the equation: y = 1 4 x + 2 .


The graph shows the x y-coordinate plane. The x and y-axis each run from -12 to 12. A line passes through the points “ordered pair 0, 2” and “ordered pair -12, 0”.

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So far, all the equations we graphed had y given in terms of x . Now we’ll graph an equation with x and y on the same side.

Graph the equation x + y = 5 .

Solution

Find three points that are solutions to the equation. Remember, you can start with any value of x or y .

The figure shows three algebraic substitutions into an equation. The first substitution is for x = 0, with 0 shown in blue. The next line is x + y = 5. The next line is 0, shown in blue + y = 5. The next line is y = 5. The last line is “ordered pair 0, 5”. The second substitution is for x = 1, with 1 shown in blue. The next line is x + y = 5. The next line is 1, shown in blue + y = 5. The next line is y = 4. The last line is “ordered pair 1, 4”. The third substitution is for x = 4, with 4 shown in blue. The next line is x + y = 5. The next line is 4, shown in blue + y = 5. The next line is y = 1. The last line is “ordered pair 4, 1”.

We list the points in a table.

x + y = 5
x y ( x , y )
0 5 ( 0 , 5 )
1 4 ( 1 , 4 )
4 1 ( 4 , 1 )

Then plot the points, check that they line up, and draw the line.

The graph shows the x y-coordinate plane. The x and y-axis each run from -7 to 7. A line passes through three labeled points, “ordered pair 0, 5”, “ordered pair 1, 4”, and ordered pair 4, 1”. The line is labeled x + y = 5.
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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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im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
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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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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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what is the problem that i will help you to self with?
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what's the easiest and fastest way to the synthesize AgNP?
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China
Cied
types of nano material
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
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what is the function of carbon nanotubes?
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what is nanomaterials​ and their applications of sensors.
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what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
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
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
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how do you translate this in Algebraic Expressions
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Prealgebra. OpenStax CNX. Jul 15, 2016 Download for free at http://legacy.cnx.org/content/col11756/1.9
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