First graph each equation as explained in my previous post but only shade lightly. Next determine if the problem is asking for the intersection (and) or union (or).
If it is an intersection then the solution for the system of two or more equations is only where all of the graphs overlap.
If it is a union it is where any of the shading is located.
Be careful of infinite solutions and no solutions.
For example: Graph the equations y > 3x - 2 and y < x + 2.
First graph y > 3x - 2.
Since, this is a greater than equation like to make a little arrow pointing up indicating which side of the line will be shaded.
Now graph y < x + 2
Again, I like to make little arrow indicating which side of the line the shading is on.
It is easy to get confused when graphing equations with one variable. The first thing that you need to determine is if the question is asking for the equation to be graphed on a number line or in the coordinate plane. Lets take the example mentioned in a comment on the previous post How to Graphing Inequalities in the Coordinate Plane.
x > 6 and x > 2.
First we need to recognize the inequality that answers the problem. Any x value that will correctly solve x > 6 will also work for x > 2. This means that x > 6 is a subset of x > 2. Since this is an and problem that is a solution has to work for both inequalities to be part of the solution set. This makes x > 6 the answer since it everything that works for x > 6 will also work for x > 2.
With that said we now need to determine which type of graph we need to create, assuming that you have to graph your answer. Note that in the graphs below each grid line represents 2.
If you are asked to graph the inequality on a number line your answer would be the red line, the one on top, of the graph below. I've drawn both lines so that you can see everywhere the red line is the orange is also.
Remember to make your left end points empty since the inequality is greater than but not equal to.
If you are asked to graph the inequality in the coordinate plane your answer should look like the yellowish/greenish area on the right side of the yellow line in the graph below. Like the graph above everywhere the yellow area is the green area is also.
Remember to dash your lines since this a greater than but not equal to.
Now if the problem as or instead of and you would the green and orange lines would be your answer since your solutions in the solution set would only have to answer one of the equations. Everything that works in x > 6 also works for x > 2.
Find the solution of a single variable inequality equation.
Assumptions:
The at you know how find the value of an inequality equation with a single variable.
Concepts:
Write the problem as two variations to remove the absolute value.
Solve as normal.
Directions: Since absolute values can contain either a positive or negative number that is always evaluated to a positive number. We need to write the equation twice since we don't know if it is positive or negative. Let's use the example: | 9 + x | < 7
| 9 + x | < 7
(9 + x ) < 7 9 - 9 + x > 7 - 9 x > -2
-(9 + x ) < 7 -(9 + x )/-1 < 7/-1 (9 + x ) > -7 9 + x > -7 9 - 9 + x > -7 - 9 x > -16
x < -2 and x > -16 -16 < x > -2
Many text books have students memorize the fact that you change the inequality sign and make the right side negative and positive but I have found that the method shown here helps students to make fewer mistakes.
Things to remember when solving absolute value inequalities.
Isolate the absolute value to one side of the equation first.
Remember to change the inequality sign when multiplying or dividing by a negative number.
Less than (<) are usually and statements.
Greater than (>) are usually or statements.
Watch out for the exceptions such as |x| < 0 or any other value less than 0, since absolute values always evaluate to be positive it will never be less than 0.
Watch out for the exception |x| > -1 which is all values of x. Again since the absolute value always evaluates to be positive any number that you insert will always work.
And statements may be written two different ways whereas or statements may only be written one way.
Having taught Algebra 1 for over 8 years I have always felt that there is a connection between how well a student understands arithmetic and how well they do in the my Algebra I class. So I decided to do a little bit of research to determine if this could be supported. Here is a graph of the results. The blue line through the graph is a regression line.Details about the study:
2 classes
11 students: 10 & 11 graders
14 students: 9 graders and one 12 grader
Arithmetic test: 1 minute timed test of multiplying numbers 1 through 10.
Accuracy of test student responses was mostly 90-100%
100 multiplication problems were provided
Highest number problems completed 57.
Highest number of problems completed correctly 56.
Grade came from the previous quarter to determine the level of success in Algebra I because it was available for all students.
Trends and Interpretation: Except for the two outliers indicated as yellow the trend seems to say that the better you are at arithmetic the better your score will be in Algebra I. The outliers can be explained as a student that is unmotivated and the other as a student that is trying but struggling. Though this isn't enough data to make a case it sure does begin to show a pattern that supports my hypothesis. I'm not sure how good of a predictor it is because as you move along any grid line there is quite a spread between the scores. Also there is the variable of different grading style between 2 teachers for the students grades. It would be great if someone would duplicate this study with the same teacher and a larger student populous. It should be noted that you don't see any grid points in the upper left and the lower right indicating a good arithmetic score and a bad grade or a good grade and a bad arithmetic score. Thus I feel that the claim can be supported that the better you are at arithmetic the better you will do in Algebra I. Please feel free to add you data in the comments below. You can find the Math Facts test at worksheetshare.com.
Is it late and you've run out of graph paper for that assignment that is due tomorrow. Have no fear free graph paper is here. Check out this slick little website printfreegraphpaper.com where you can select your grid size and make your own graph paper.
The goal of solving systems of equations is to determine if two or more lines intersect and if they do where do they intersect. This lesson will only deal with two lines.
First we need to determine how the lines of the two equations relate to each other. Do they intersect, are they parallel, or are they the same line. The mathematical ways of describing this is are they consistent, inconsistent, dependent, and independent. There are three options as shown in the picture below: To find the answer to the system you need to ask one or two questions depending on the problem.
Intersecting lines: Independent and consistent
Parallel lines: Independent and inconsistent
Same line: Inconsistent
Question 1: First write the equation in slope intercept form. Then ask the question is the slope in each equation the same or different? If it is the same then they are parallel and you need to proceed to the next question. If it is different then the point at which the lines cross is the solution to the system and we are done.
Question 2: Since the slope is the same the lines are parallel. Now you need to ask the question "Are they the same line?" If the y-intercepts are the same then they are the same line and there is an infinite number of solutions. The line is the solution to the system of equations because it represents all of the points that work in both equations which are really the same equation just written differently.
If they are not the same then there is no solution, that is the lines do not cross.
Practice Problems:
Find the solution of the equations y = 2x + 4 and y = 0.5x - 2
First we recognize that the equations are in slope intercept form (y = mx + b) and that the slopes of each of the lines are different so we know that they will cross at some point. The next step is to graph the line. It is very important that your lines be accurate so I would recommend placing as many calculated points as possible on the graph for each line.
The solution is ( -4, -4 ). Make sure to check your answer by plugging the point into both problems. y = 2x + 4 -4 = 2 ( -4) + 4 -4 = -8 + 4 -4 = -4
and y = 0.5x - 2 -4 = 0.5 (-4) - 2 -4 = -2 - 2 -4 = 4
Lets try another example: Solve the system of equations: y= 4 2x - 3y = -6
First off notice that the second equation is in standard form and not slope intercept form. We need to rearrange it so that we can compare the slopes. 2x - 3y = -6 Subract 2x from both sides. -3y = -2x - 6 Now divide both sides by -3 y = (2/3)x + 2
Again notice that the slopes are not the same so we know that they will cross at some point. Graph your line, making sure to plot lots of points to keep it accurate.
The solution is ( 3, 4). Check your answer y = 4 4 = 4. Remember that this equation doesn't care about the x value so it can be anything.
2x - 3y = -6. Always use the original problem in case you made a mistake when you were changing the problem to slope-intercept form. 2(3) - 3(4) = -6 6 - 12 = -6 -6 = -6
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.
The need arose to be able to quickly graph points and lines during class presentations so I decided to develop the following flash application. The curve is a great addition but I don't know that it will be as useful. It's pretty simple to use just select the point, line, or curved button. Next put them on the graph. If you mess up just hit the red X and it all goes away. Below is a screen shot of the program. Just give it a click to head on over to my other website where you can give it a try. Feel free to come back and leave me some comments on how useful you thing this is.
I've created a numeric question generator for Moodle that will create MoodleXML files that can be imported into your course. It will create addition, subtraction, multiplication, division, powers, and square root problems based on the upper and lower bounds of numbers that you give it. It can also limit the answers to only positive numbers and/or only integer answers. Check it out here: http://aschool.us/moodle-scripts/math-questions.php
Here is a little flash file that I created this morning to help you understand how to find the greatest common factor between two numbers. I hope to add audio to this later but don't have the time at the moment. Simply click the button at the bottom to work your way through the process. To find the greatest common factor between 18 and 24 first you need to find the prime factorization of each. To do this divide 18 by 2 and you get 9. Nine is not a prime, it can be divided by 3. So the prime factorization of 18 is 2x3x3. Next do the same with 24. Twenty-four divided by 4 is 6. Neither 4 or 6 are prime so they need to be broken down farther. Four is divisible by 2 and 6 is divisible by 2 and 3. Making the prime factorization of 24 = 2x2x2x3. Next find the common numbers that are in each prime factorization of each number. Both 18 and 24 contain a 2 and a 3. Multiply these together to get 6. Thus 6 is the greatest common factor of 18 and 24. To download download the file flash file can be downloaded from to www.woehler.us
I've created a set of resouces that can help you create graphing materials. The were created using Inkscape a open-source vector drawing program. The files are saved as svg, pdf, gif, and png formats. You will find a ready made set of graph paper and individual graphs that can be inserted into word, open office, a web page or where ever you want. I've also provided the original svg files so that you can quickly make any changes that you want.
You can find them on my website at www.woehler.us. Let me know if you find these to be useful.
While working with my Algebra 1 students to try and teach them how to graph lines. I came up with the idea to create a flash module that would work with moodle to record student's graphs. This idea came out of my previous project which was to create a quick way to draw graphs while lecturing.
This module is pretty slick you can use it to teach students how to graph linear inequalities and equality graphs. My other web page explains the details but here are the basics.
Select your line type, which is by default solid for equality next click to points that are on the line. The flash module extends the line to the edge of the graph. If you are going to graph an inequality then click on shading. Click to points on the line and then which side of the line should be shaded.
You will notice that when you select shading you now have the option of two different line types. The dashed line is for inequality graphs.
Here is the really cool part though. When you create the graph, if you set a value in the flashvars variable the equation of the line can be recorded to a form value in the web page which can then be submitted for grading.
I've implemented this version with moodle but I'll post that information later since I'm not using the latest version of moodle and want to cleaned up the code a little bit and make sure that it is compatible with the latest version.
Well I finally finished my graphing utility. I'm sure there are many cool features that I could add but I don't have time for that at the moment. Please let me know what you think. I've posted a copy at www2.woehler.us/cms/. You can find the flash/exe/hqx version and instructions there.
I hope that this will be a helpful resource to math teachers presenting graphing concepts such as slope intercept form, y-intercept form, standard form, and point slope form or whatever other type of graph you are planning on using. It even works with inequalities.
My algebra students are stuggling with the concept of graphing so I've created a couple of Flash modules that facilitate teaching graphs. One of them you can simply put points on the graph by clicking and it will tell you the coordinates on the graph. The other one allows you to click two points on the graph to draw a line segment and change the line segment into an equality or inequality graph.
This post is just to see if there is some interest in these modules. I'm looking for some math teachers out there that might be interesting in testing it out for me and giving me some feedback before I post it for free on the internet.
You can put this module into a powerpoint, show it in a web page, or run it as a program.
Please post a comment if this sounds interesting to you.
