- Graph inequalities in a xy coordinate graph.
Assumptions:
- Ability to graph a line using the slope-intercept form (y = mx + b)
Concepts:
- The shaded area of a graph represents all of the coordinates that will work in a given equation.
- A solid edge of the shaded area means that the edge is part of the solutions to the equation.
- A dashed edge of the shaded area means that the edge of the graph is not part of the solutions.
Directions:
Graph the equation
Step 1: Draw the graph just as you would y = x . This equations in slope intercept form would look like this
. The 0 means that you will go through the origin, place a point there. Now use the slope to draw the rest of the line. From the origin go up one and to the right one and place another point. Repeat until you have several points. 
Now draw a solid line because the equation to be graphed is greater than or equal to. Your graph should now look like this:
Step 2: Next shade everywhere above the line because the equation states that the y values are greater than or equal to the line for any given x value.
Now check your answer by inserting a couple of points from the shaded area and non-shaded area.
Shaded
Does the point ( 1, 2) work in the equation? yes 
Does the point ( -1, 0) work in the equation? yes
Non-shaded
Does the point ( 1, 0) work in the equation? no 
Does the point ( 2, 1) work in the equation? no
Lets try another one.
Graph graph y > 2x + 3
Remember the steps: plot some points, draw the line (solid if equal to, dashed if greater than or less than), shade above with greater than, shade below with less than.
The line will cross the y axis as 3 then go up 2 and over 1 for the slope. Start by placing a point at 3 on the y axis. Next use the slope to place 2 more dots, then make a dashed line through the dots.
The equation uses the greater than inequality so it should be shaded above the line.
Now that we have the common ones out of the way lets look at the ones that may trip you up such as the ones with only one variable like y > 2 and x < -3.
Graph y > 2
Remember that is just a horizontal line. This is just a horizontal line that is shaded above the line and dashed because it is not equal to the line it is only greater than the line.
Graph x < -3
Remember that is just a vertical line. This is just a vertical line that is shaded to the left of the line and dashed because it is not equal to the line it is only less than the line. The x values on the left are less than the line.
Things to remember when graphing inequalities:
≥ Solid line and shaded above the line.
≤ Solid line and shaded below the line
> Dashed line and shaded above the line
y > # Horizontal line and shaded above the line
y < # Horizontal line and shaded below the line
x > # Vertical line and shaded on the right side of the line
x < # Vertical line and shaded on the left side of the line.
46 comments:
How would you graph 2 < x >6 and 2x+3 > 5 together?
First I'm not sure what you mean by 2 < x > 6? Do you mean that x > 2 and x > 6 or do you mean x > 2 and x < 6?
In the first option x > 6 is a subset of x > 2 so graphing x > 2 would cover both of them. To graph x > 2 draw a dashed vertical line through 2 on the x axis and shade the right side of it. Next graph a dashed line for 2x + 3 > 5 and shade above the line. Shade it darker where both shaded areas cross.
If you meant that 2 < x and x < 6 then draw two dashed vertical lines through 2 and 6 on the x axis and shade between them. Then graph the other line as before.
it is just another way of saying x>2 and x>6 instead of writing it out into a longer virsion you can put them togeather
In response to Anonymous since x > 2 and and x > 6. If you graph these you will notice that only when the input for the x > 2 problem is greater than 6 will your input correctly solve both problems. Since this is an and problem the solution set must work in both inequalities. That being the case then it makes sense that correct way to write the answer would be x > 6.
when you do the problems like that aaron, you would put it on a line graph. you would plot the numbers and then since x is bigger than 2 and bigger than 6 your arrow would be moving all the way tword the 6 and beyond.
In reply to the previous post about a line graph. Yes, you could graph it on a line graph but some times it needs to be graphed in the coordinate plane when you have a system of equations, such as graphing the intersection of two equations:
y > 2x + 2
x < 10
How would I graph 1>2x+5
To graph 1>2x+5 first solve for x by subtracting 5 from both sides and then dividing by 2. This leaves you with -2 > x which can be rewritten as x < -2.
To graph on the number line draw a circle on the -2 and an arrow to the left. On a 2-dimensional graph draw a vertical dashed line through the -2 on the x-axis and shade the left side of the line all the way to the left edge of the graph.
There are 2 errors in the page I ran into. Under 'Directions', in part 2 under the first example of shading:
"Does (1,2) work in the equation? yes 1>2" should read as 2>1
"Does the point(2,1) work in the equation? no 2>1" should read as 1>2 (assuming you don't have a way to denote 'not greater than or equal to' on the webpage)
cheers!
How would I graph x>y-1?
Thank you for noting those errors they have been fixed.
To graph x > y - 1 first add one to both sides, giving you x + 1 > y which could be rewritten in slope-intercept for or y-intercept form as y < x + 1. This would be shaded below a dashed line.
26 is an "and" problem. You can only graph it on a number line(x>2 and x>6).2x+3>5 is basically an equation except it uses an inequality. 2x+3>5
-3-3
2x>2
divide2 2
x>1
Then you graph it on a coordinate plane. To graph the two together is impossible(I think)
In response to the discussion I've created a new post Graphing Inequalities With One Variable. Hopefully this will help clarify some of the discussion.
How would I graph y < -x?
Would the slope be a negative 1 over a negative one, or would it be a negative 1 over 1?
Your second answer would be correct. It can be written three different ways:
-(1/1), -1/1, or 1/(-1) which all are equivalent. Next draw your dashed line and shade below it.
How would I graph y < -x?
It may be easier to graph y < -x if it is written this way
y < (-1/1)x
Here the slope is (-1)/1 meaning that from the origin you go down one and to the right one. We can write it this way since (-1)/1 = 1/(-1) = -(1/1) = -1. You could also go up one and back one. This would give you the three points to make your graph with: (-1, 1), (0, 0), and (1, -1). Remember to make a dashed line and shade below the line.
Is it possible to graph something like:
-3w-5> 10
Check out the other blog post that I have posted since this one has raised so many questions about graphing inequalities with one variable. Since you are using one variable I'm assuming that you are graphing on a number line instead of a coordinate plane.
i am having the biggest meltdown over this because i do not understand. please can someone explain how i could graph and work out y<-x+5
please i need help as soon as possible
thank you very much in advance :)
To graph y<-x+5 you will need to start by going up to the point (0,5), on the y-axis. Next find your slope which is -1. This is equivalent to (-1)/1 so from that point go down one and to the right one to the point (1, 4) do it again until you have enough points to make your line. Since it is < the line should be dashed and shaded below the line. The following points should be included as part of your shading: (0, 0), (-5, -5), (-5, 2), and (3, -5).
how i graph -3x+4y>6 ?
how would you graph y= -|x=2|-1
how would i graph x-3y>-3
Whatg happen if I have two inequalities and I'm trying to find the common points between them? For example, y > or equal to x and y<x+1. Thanks
The easiest way to graph -3x+4y>6 would be to find the x and y intercepts, by setting each value to zero and solving for the other value. To find the y-intercept set x to zero which gives you -3*0+4y>6 simplifying to 4y>6 making y>3/2 or y>1.5. Do the same for the x-intercept set y equal to zero and solve for x. Once you have the two points draw a dashed line in between and shade above the line.
To graph inequalities with two variables graph as you would with one inequality but only shade where both inequalities are shaded. The shaded area should look more like a v or an area shaded between two lines. Also be careful for non existent solutions or overlapping solutions.
How would you graph 2(y+1) > 5x? I am not sure what to do since I cant set 2(y+1) equal to 5x. I have tried dividing out the 2 but cant graph y+1>2.5x since I dont have a y intercept. Would I move the 1 over to the other side then use that as the y intercept?
Thanks,
Anon
For the equation 2(y+1) > 5x you are correct to divide both sides by 2 then you will need to subtract 1 from both sides giving you a final answer of y > 2.5x - 1. You y-int will be -1. I prefer to write the slope as an improper fraction rather than a decimal. I would leave it as 5/2 showing a rise of 5 and a run of 2 from the y-intercept at (0, -1).
Is it possible to graph -2y>x as it is? Or would you have to divide the -2 and the x? and get y>-1/2x?
Chris. Your answer is almost correct. Remember to change the inequality whey dividing by a negative number. Your answer should be
y < (-1/2)x
When the equation is in a y-intercept form it is easier to graph.
I've added a new post that demonstrates how to graph two inequalities.
Hello there!
You see... I'm having trouble with graphing an inequality...
y ≥ x - 1
So, I already placed -1 just below the point of origin and followed a 1/1 slope. So, yes, I got the line. I'm confused with the shading. I solved for the point within the line, above the line and below the line. It seems that they're all solutions to the equation.
ex: On the line: (-1, -2)
y ≥ x - 1
-2 ≥ -1 -1
-2 ≥ -2 = True
Above the line: (2,3)
y ≥ x - 1
3 ≥ 2 - 1
3 ≥ 1 = True
Below the line: (-3,-2)
y ≥ x - 1
-2 ≥ -3 -4
-2 ≥ -7 = True
Am I doing something wrong? Please help...
You answers are correct. If you were to try the point (-2, -3) that would fail the test because it would be below the line.
Use the flash app at the bottom of the page to experiment with what I'm saying. I think you will see it. Click (0,-1) then click (1,0) then press the >= button.
http://aschool-us.blogspot.com/2011/06/graphing-multiple-inequalities-using.html
how would you graph -2x+3y>-6
how would you graph 2x-7y< -14
Nickky420: To graph 2x-7y< -14 I would move the 2x to the other side by subtracting 2x from both sizeds giving your -7y < -2x - 14. Next divide both sides by -7 resulting in an answer of y > (2/7)x + 2. Draw a dashed line from (0,2) with a slope of 2/7. Shade above the line.
How would you graph 3x+2y=6
Sorry about the late reply this one slipped between all of the other emails. To graph 3x+2y=6 you first need to change it into y-intercept or slope-intercept form: y = m*x + b. Then you can graph it using the directions found at the beginning of the Graphing Systems of Equations since it is not an inequality but an equality.
How would i graph 5x+10y<-30
To graph 5x + 10y < -30 first put it in y-intercept form by subtracting 5x from both sides and then dividing everything by 10.
Dear Aaron, how would you graph x (Greater then or equal to) 4?
I think that I would shade the right side of the line, right?
And btw, I think you're a great role model. You wrote a great article, and you help tons of people out. That's something to be proud of!
You are correct. The right side should be shaded. Also make sure that the line is dashed since you didn't say it was equal.
I'm glad so many people have shown an interest in it and found it useful.
How would you graph -1<n≤5 on a coordinate plane?
To graph -1 < n ≤5 you will need to draw a dashed line at -1, a solid line at 5 and shade everything in between them. Since n is not a standard axis such as x or y I'm not sure if your lines will be vertical or horizontal.
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