The document discusses solving systems of linear equations with two or three variables. For two variables, the solutions can be one point (consistent, one solution), all points on a line (consistent, dependent with infinite solutions), or no solution (inconsistent). For three variables, the solutions can be one point, all points on a line, or no solution, which are determined by whether the graphs (planes) intersect at a point, line, or not at all. The document demonstrates using elimination methods to solve sample systems of two and three variable equations algebraically.
Identify basic properties of equations
Solve linear equations
Identify identities, conditional equations, and contradictions
Solve for a specific variable (literal equations)
Identify basic properties of equations
Solve linear equations
Identify identities, conditional equations, and contradictions
Solve for a specific variable (literal equations)
AN EQUATION WHICH CAN BE WRITTEN IN THE FORM OF ax+by+c=0 WHERE a,b and c ARE REAL NUMBERS.
YOU WILL GET TO KNOW HOW TO REPRESENT THE EQUATIONS IN A GRAPH.
AN EQUATION WHICH CAN BE WRITTEN IN THE FORM OF ax+by+c=0 WHERE a,b and c ARE REAL NUMBERS.
YOU WILL GET TO KNOW HOW TO REPRESENT THE EQUATIONS IN A GRAPH.
1 Part 2 Systems of Equations Which Do Not Have A Uni.docxeugeniadean34240
1
Part 2: Systems of Equations Which Do Not Have A Unique
Solution
On the previous pages we learned how to solve systems of equations using Gaussian
elimination. In each of the examples and exercises of part 1(except for exercise 1 parts d and e)
the systems of equations had a unique solution. That is, a single value for each of the variables.
In example 3 we found the solution to be 7 23 3, . This means that the graphs of the two lines in
example 3 intersect at this unique point. In 2-space, the xy-plane, we have the geometric bonus
of being able to draw a picture of the solutions to a system of two equations two unknowns.
Clearly, if we were asked to draw the graphs of two lines in the xy-plane we have 3 basic
choices/cases:
1. Draw the two lines so they intersect. This point of intersection can only happen once for
a given pair of lines. That is, the two lines intersect in a unique point. There is a unique
common solution to the system of equations. Discussed in part 1.
2. Draw the two lines so that one is on "top of" the other. In this case there are an infinite
number of common points, an infinite number of solutions to the given system. Discussed
in part 2.
3. Draw two parallel lines. In this case there are no points common to both lines. There is
no solution to the system of equations that describe the lines. Discussed in part 2.
The 3 cases above apply to any system of equations.
Theorem 1. For any system of m equations with n unknowns (m < n) one of the following cases
applies:
1. There is a unique solution to the system.
2. There is an infinite number of solutions to the system.
3. There are no solutions to the system.
Again, in this section of the notes we will illustrate cases 2 and 3. To solve systems of
equations where these cases apply we use the matrix procedure developed previously.
Example 6. Solve the system
x + 2y = 1
2x + 4y = 2
2
It is probably already clear to the reader that the second equation is really the first in
disguise. (Simply divide both sides of the second equation by 2 to obtain the first). So if we
were to draw the graph of both we would obtain the same line, hence have an infinite number of
points common to both lines, an infinite number of solutions. However it would be helpful in
solving other systems where the solutions may not be so apparent to do the problem
algebraically, using matrices. The matrix of the system with its simplification follows. Recall,
we try to express the matrix
1 2 1
2 4 2
in the form 1
2
1 0
0 1
b
b
from which we can read off the
solution. However after one step we note that
1 2 1
2 4 2
1 22 R R
1 2 1
0 0 0
. It should be clear to the reader that no matter what further
elementary row operations we perform on the matrix
1 2 1
0 0 0
we cannot change it to the form
we hoped for, namel.
A rate of change is a rate that describes how one quantity changes in relation to another.
A constant rate of change is the rate of change of a linear relationship.
Essential Question How can you find the unit rate from a line on a graph that goes through the origin?
Objective understand slope as it relates to rate of change
Essential Question How can you find the unit rate from a line on a graph that goes through the origin?
Objective understand slope as it relates to rate of change
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
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Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdfTechSoup
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
1. Systems of Linear Equations The solution will be one of three cases: 1. Exactly one solution, an ordered pair (x, y) 2. A dependent system with infinitely many solutions 3. No solution Two Equations Containing Two Variables The first two cases are called consistent since there are solutions. The last case is called inconsistent.
2. With two equations and two variables, the graphs are lines and the solution (if there is one) is where the lines intersect. Let’s look at different possibilities. Case 1: The lines intersect at a point ( x , y ), the solution. Case 2: The lines coincide and there are infinitely many solutions (all points on the line). Case 3: The lines are parallel so there is no solution. dependent consistent consistent independent inconsistent
3. If we have two equations and variables and we want to solve them, graphing would not be very accurate so we will solve algebraically for exact solutions. We'll look at two methods. The first is solving by substitution. The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation. Let's solve for y in the second equation. You can pick either variable and either equation, but go for the easiest one (getting the y alone in the second equation will not involve fractions). Substitute this for y in the first equation. Now we only have the x variable and we solve for it. Substitute this for x in one of the equations to find y. Easiest here since we already solved for y .
4. So our solution is This means that the two lines intersect at this point. Let's look at the graph. We can check this by subbing in these values for x and y . They should make each equation true. Yes! Both equations are satisfied.
5. Now let's look at the second method, called the method of elimination. The idea is to multiply one or both equations by a constant (or constants) so that when you add the two equations together, one of the variables is eliminated. Let's multiply the bottom equation by 3. This way we can eliminate y's. (we could instead have multiplied the top equation by -2 and eliminated the x's) Add first equation to this. The y 's are eliminated. Substitute this for x in one of the equations to find y. 3 3
6. So we arrived at the same answer with either method, but which method should you use? It depends on the problem. If substitution would involve messy fractions, it is generally easier to use the elimination method. However, if one variable is already or easily solved for, substitution is generally quicker. With either method, we may end up with a surprise. Let's see what this means. Let's multiply the second equation by 3 and add to the first equation to eliminate the x 's. 3 3 Everything ended up eliminated. This tells us the equations are dependent. This is Case 2 where the lines coincide and all points on the line are solutions.
7. Now to get a solution, you chose any real number for y and x depends on that choice. If y is 0, x is -5 so the point (-5, 0) is a solution to both equations. If y is 2, x is -1 so the point (-1, 2) is a solution to both equations. So we list the solution as: Let's solve the second equation for x . (Solving for x in either equation will give the same result) Any point on this line is a solution
8. Let's try another one: Let's multiply the first equation by 2 and add to the second to eliminate the x 's. 2 2 This time the y 's were eliminated too but the constants were not. We get a false statement. This tells us the system of equations is inconsistent and there is not an x and y that make both equations true. This is Case 3, no solution. There are no points of intersection
9. Systems of Linear Equations The solution will be one of three cases: 1. Exactly one solution, an ordered triple (x, y, z) 2. A dependent system with infinitely many solutions 3. No solution Three Equations Containing Three Variables As before, the first two cases are called consistent since there are solutions. The last case is called inconsistent.
10. Planes intersect at a point: consistent with one solution With two equations and two variables, the graphs were lines and the solution (if there was one) was where the lines intersected. Graphs of three variable equations are planes. Let’s look at different possibilities. Remember the solution would be where all three planes all intersect.
11. Planes intersect in a line: consistent system called dependent with an infinite number of solutions
14. We will be doing elimination to solve these systems. It’s like elimination that you learned with two equations and variables but it’s now the problem is bigger. Your first strategy would be to choose one equation to keep that has all 3 variables, but then use that equation to “eliminate” a variable out of the other two. I’m going to choose the last equation to “keep” because it has just x . coefficient is a 1 here so it will be easy to work with
15. keep over here for later use Now use the third equation multiplied through by whatever it takes to eliminate the x term from the first equation and add these two equations together. In this case when added to eliminate the x’s you’d need a -2. put this equation up with the other one we kept
16. We won’t “keep” this equation, but we’ll use it together with the one we “kept” with y and z in it to eliminate the y ’s. Now use the third equation multiplied through by whatever it takes to eliminate the x term from the second equation and add these two equations together. In this case when added to eliminate the x’s you’d need a 3.
17. So we’ll now eliminate y ’s from the 2 equations in y and z that we’ve obtained by multiplying the first by 4 and the second by 5 keep over here for later use We can add this to the one’s we’ve kept up in the corner
18. Now we are ready to take the equations in the corner and “back substitute” using the equation at the bottom and substituting it into the equation above to find y . keep over here for later use Now we know both y and z we can sub them in the first equation and find x These planes intersect at a point, namely the point (2, -1, 1). The equations then have this unique solution. This is the ONLY x , y and z that make all 3 equations true.
19. Let’s do another one: we’ll keep this one Now we’ll use the 2 equations we have with y and z to eliminate the y ’s. If we multiply the equation we kept by 2 and add it to the first equation we can eliminate x ’s. I’m going to “keep” this one since it will be easy to use to eliminate x’s from others. If we multiply the equation we kept by 3 and add it to the last equation we can eliminate x ’s.
20. we’ll multiply the first equation by -2 and add these together Oops---we eliminated the y ’s alright but the z ’s ended up being eliminated too and we got a false equation. This means the equations are inconsistent and have no solution. The planes don’t have a common intersection and there is not any ( x , y , z ) that make all 3 equations true.
21. Let’s do another one: we’ll keep this one If we multiply the equation we kept by -2 and add it to the second equation we can eliminate x ’s. Now we’ll use the 2 equations we have with y and z to eliminate the y ’s. If we multiply the equation we kept by -4 and add it to the last equation we can eliminate x ’s. let's “keep” this one since it will be easy to use to eliminate x ’s from others.
22. multiply the first equation by -1 and add the equations to eliminate y . Oops---we eliminated the y ’s alright but the z ’s ended up being eliminated too but this time we got a true equation. This means the equations are consistent and have infinitely many solutions. The planes intersect in a line. To find points on the line we can solve the 2 equations we saved for x and y in terms of z .
23. First we just put z = z since it can be any real number. Now solve for y in terms of z . Now sub it - z for y in first equation and solve for x in terms of z . The solution is (1 - z , - z , z ) where z is any real number. For example: Let z be 1. Then (0, -1, 1) would be a solution. Notice is works in all 3 equations. But so would the point you get when z = 2 or 3 or any other real number so there are infinitely many solutions.