Mathematics
Solving Linear Equations
Solving linear equations involves finding the value of the variable that makes the equation true. This is typically done by performing operations to isolate the variable on one side of the equation. The goal is to simplify the equation to the form "variable = value" or "value = value," allowing for the solution to be easily determined.
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9 Key excerpts on "Solving Linear Equations"
eBook - PDF- (Author)
- 2015(Publication Date)
- For Dummies(Publisher)
But these would come next in the reverse order of operations. When Solving Linear Equations, the goal is to isolate the variable you’re trying to find the value of. Isolating it, or getting it all alone on one side, can take one step or many steps. And it has to be done according to the rules — you can’t just move things willy‐nilly, helter‐skelter, hocus pocus . . . you get the idea. Solving Equations with Two Terms Linear equations contain variables raised to the first power. The easiest types of linear equations to solve are those consisting of just two terms. The following equations are all examples of linear equations in two terms: 14x = 84 −64 = 8y 9 5 18 z = 7 6 35 9 w = Linear equations that contain just two terms are solved with multiplication, division, reciprocals or some combinations of the operations. Devising a method using division One of the most basic methods for solving equations is to divide each side of the equation by the same number. Many formulas and equations include a coefficient (multiplier) with the variable. To get rid of the coefficient and solve the equation, you divide. The following example takes you step by step through solving with division. 226 Part III: Solving Algebraic Equations Solve for x in 20x = 170. 1. Determine the coefficient of the variable and divide both sides by it. Because the equation involves multiplying by 20, undo the multiplication in the equation by doing the opposite, which is division. Divide each side by 20: 20 20 170 20 x = 2. Reduce both sides of the equal sign. 20 20 170 20 x = x = 8.5 Do unto one side of the equation what the other side has had done unto it. Next, I show you two examples with practical applications embedded in them. Your teacher needs to buy 300 doughnuts for a big parent meeting. How many dozen doughnuts is that? Let d represent the number of dozen doughnuts you need. A dozen has 12 doughnuts, so 12d = 300. Twelve times the number of doughnuts you need has to equal 300.
eBook - PDFIntroductory Mathematics
Concepts with Applications
- Charles P. McKeague(Author)
- 2013(Publication Date)
- XYZ Textbooks(Publisher)
©iStockphoto.com/sculpies HOW TO Steps to Solve a Linear Equation in One Variable Step 1: Simplify each side of the equation as much as possible. This step is done using the commutative, associative, and distributive properties. Step 2: Use the addition property of equality to get all variable terms on one side of the equation and all constant terms on the other, then combine like terms. A variable term is any term that contains the variable. A constant term is any term that contains only a number. Step 3: Use the multiplication property of equality to get the variable by itself on one side of the equation. Step 4: Check your solution in the original equation if you think it is necessary. Note Once you have some practice at solving equations, these steps will seem almost automatic. Until that time, it is a good idea to pay close attention to these steps. OBJECTIVES Chapter 11 Solving Equations 558 EXAMPLE 1 Solve: 3(x + 2) = −9. Solution We begin by applying the distributive property to the left side. Step 1 3(x + 2) = −9 3x + 6 = −9 Distributive property Step 2 3x + 6 + (−6) = −9 + (−6) Add −6 to both sides. 3x = −15 Step 3 3x __ 3 = −15 ____ 3 Divide both sides by 3. x = −5 This general method of Solving Linear Equations involves using the two properties we have already developed. We can add any number to both sides of an equation or multiply (or divide) both sides by the same nonzero number and always be sure we have not changed the solution to the equation. The equations may change in form, but the solution to the equation stays the same. Looking back to Example 1, we can see that each equation looks a little different from the preceding one. What is interesting, and useful, is that each of the equations says the same thing about x. They all say that x is −5. The last equation, of course, is the easiest to read. That is why our goal is to end up with x isolated on one side of the equation.
eBook - PDFEssential Maths
for Business and Management
- Clare Morris(Author)
- 2007(Publication Date)
- Bloomsbury Academic(Publisher)
77 Solving equations CHAPTER 5 Intended learning outcomes By the end of your work on this chapter you should: ɀ understand what is meant by an equation and its solution ɀ be able to recognise and solve linear equations ɀ be able to recognise and solve quadratic equations using the formula method ɀ be able to solve two simultaneous linear equations ɀ feel confident in manipulating formulae ɀ understand inequalities and be able to manipulate them correctly ɀ be able to build equations to represent practical situations, and to interpret their solutions appropriately. Prerequisites Before you start work on this chapter, you should be confident in your understanding of the material of Chapters 3 and 4. 5.1 78 Essential Maths Equations in practice In Chapter 4, we’ve seen a number of examples showing how algebraic expressions can be used to represent practical relationships. Very often, we wish to go further and, rather than just looking at the general relationship, ask what value a variable will have in partic-ular circumstances. Here are two examples, based on problems we’ve already come across in the preceding chapter. Example 1: How much should we charge? In Section 4.3, we developed the relationship n = 520 – 4 p which linked the price in pence of a packet of biscuits with the number of packets sold per week. If the retailer decides that she wishes to sell 200 packets a week, how much should she charge per packet? When 200 packets are sold, n = 200, and so we can replace n in the general relationship by 200, to get 200 = 520 – 4 p . This is an equation which, as we will see, can be solved to tell us the specific value of p required. Example 2: How long will we have to wait? Also in Section 4.3, we built up the expression n = 120 + 20 k to show how the number of visitors per week to a website, n , increases with the number of weeks it’s been operating, k .
No longer available |Learn more- Alfred Basta, Stephan DeLong, Nadine Basta, , Alfred Basta, Stephan DeLong, Nadine Basta(Authors)
- 2013(Publication Date)
- Cengage Learning EMEA(Publisher)
Solving Systems of Linear Equations Algebraically and with Matrices Chapter 5 5.1 S OLVING S YSTEMS OF L INEAR E QUATIONS BY THE S UBSTITUTION M ETHOD 5.2 S OLVING S YSTEMS OF E QUATIONS U SING THE M ETHOD OF E LIMINATION 5.3 S UBSTITUTIONS T HAT L EAD TO S YSTEMS OF L INEAR E QUATIONS 5.4 I NTRODUCTION TO M ATRICES 5.5 U SING M ATRICES TO S OLVE S YSTEMS OF L INEAR E QUATIONS t the end of Chapter 4, we encountered systems of linear equations and saw that they could be solved by determining the point(s) of intersection of the graphs of the equations. In the event that a single point of intersection occurred, that point was identified as the solution, and the system was called consistent. If the lines were parallel (and hence no intersection would occur), the system was called inconsistent, and we said there was no solution. In the rather rare in-stance where the two lines had the same graph, the system was called consistent and dependent, and the solution set consisted of all points lying on their mutual graph. That visual method of solving systems possesses in-herent weaknesses. First, a high-quality graph must be in your possession, with lines drawn that are extremely precise. Even the slightest imperfection in the graph of one or both of the lines could lead you to an incor-rect determination of the solution. Second, even with a high-quality graph in hand, if the solution point were such that its coordinates were not integer valued, ex-act identification of the solution could prove to be dif-ficult or impossible. These issues lead us to consideration of methods for solution that are immune from the ambiguity of graphical analysis, methods that are wholly algebraic and computational. Such methods are free from the requirement that a high-quality graph be in our A Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
- Kevin D. Dorfman, Prodromos Daoutidis(Authors)
- 2017(Publication Date)
- Cambridge University Press(Publisher)
2 Linear Algebraic Equations 2.1 Definition of a Linear System of Algebraic Equations In this chapter, we will discuss a myriad of methods for solving a system of linear algebraic equations on the computer. The general form of a system of linear algebraic equations is a 1,1 x 1 + a 1,2 x 2 + . . . + a 1, n x n = b 1 a 2,1 x 1 + a 2,2 x 2 + . . . + a 2, n x n = b 2 . . . . . . a n ,1 x 1 + a n ,2 x 2 + . . . + a n , n x n = b n (2.1.1) where we are trying to solve for the various unknowns x i . As shown in the above descrip-tion, we will assume that the number of equations is equal to the number of unknowns. In the context of our previous definition of mathematical models, let’s write this equation system in matrix form Ax = b (2.1.2) The coefficient matrix A = ⎡ ⎢ ⎢ ⎢ ⎣ a 1,1 a 1,2 · · · a 1, n a 2,1 a 2,2 · · · a 2, n . . . . . . . . . a n ,1 a n ,2 · · · a n , n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.3) is the operator for this problem, the unknown vector is x = ⎡ ⎢ ⎢ ⎢ ⎣ x 1 x 2 . . . x n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.4) 46 Linear Algebraic Equations and the forcing function is the constant vector b = ⎡ ⎢ ⎢ ⎢ ⎣ b 1 b 2 . . . b n ⎤ ⎥ ⎥ ⎥ ⎦ (2.1.5) You may wonder why we are devoting so much time in this book to the “simple” problem of solving this system of equations. In particular, you may remember solving numerous mass and energy balance problems that ultimately reduced to a system of linear algebraic equations and hardly felt the need to resort to a computer to get an answer. This is because the problems you were trying to solve are too small, not because linear algebraic systems are easy to solve! Consider the two problems in Fig. 2.1 . In most introductory chemical engineering classes, the problems are similar to Fig. 2.1 a. These are small problems, and the unknowns are selected such that the equations are only weakly coupled. In these small problems, you can proceed easily through the equations, solving for one variable at a time and then proceeding to the next equation.
eBook - PDF- Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis(Authors)
- 2020(Publication Date)
- Openstax(Publisher)
Read the problem. Make sure all the words and ideas are understood. Identify what we are looking for. Name what we are looking for. Choose variables to represent those quantities. Translate into a system of equations. Solve the system of equations using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence. Step 1. Step 2. Step 3. Step 4. Step 5. Step 6. Step 7. 606 Chapter 5 Systems of Linear Equations This OpenStax book is available for free at http://cnx.org/content/col31130/1.4 Step 6. Check the answer in the problem. Do these numbers make sense in the problem? We will leave this to you! Step 7. Answer the question. The numbers are 9 2 and − 9 2 . TRY IT : : 5.41 The sum of two numbers is 10. One number is 4 less than the other. Find the numbers. TRY IT : : 5.42 The sum of two number is −6. One number is 10 less than the other. Find the numbers. In the Example 5.22, we’ll use the formula for the perimeter of a rectangle, P = 2L + 2W. EXAMPLE 5.22 The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width. Solution Step 1. Read the problem. Step 2. Identify what you are looking for. We are looking for the length and width. Step 3. Name what we are looking for. Let L = the length W = the width Step 4. Translate into a system of equations. The perimeter of a rectangle is 88. 2L + 2W = P The length is five more than twice the width. The system is: Step 5. Solve the system of equations. We will use substitution since the second equation is solved for L. Substitute 2W + 5 for L in the first equation. Solve for W. Chapter 5 Systems of Linear Equations 607 Substitute W = 13 into the second equation and then solve for L. Step 6. Check the answer in the problem. Does a rectangle with length 31 and width 13 have perimeter 88? Yes. Step 7. Answer the equation. The length is 31 and the width is 13.
eBook - PDFIntroductory Algebra
Concepts with Applications
- Charles P. McKeague(Author)
- 2013(Publication Date)
- XYZ Textbooks(Publisher)
The following box shows some steps to follow when solving a system of linear equations by the elimination method. HOW TO Solving a System of Linear Equations by the Elimination Method Step 1: Decide which variable to eliminate. (In some cases one variable will be easier to eliminate than the other. With some practice you will notice which one it is.) Step 2: Use the multiplication property of equality on each equation separately to make the coefficients of the variable that is to be eliminated opposites. Step 3: Add the respective left and right sides of the system together. Step 4: Solve for the variable remaining. Step 5: Substitute the value of the variable from step 4 into an equation containing both variables and solve for the other variable. Step 6: Check your solution in both equations, if necessary. GETTING READY FOR CLASS After reading through the preceding section, respond in your own words and in complete sentences. A. How do you use the addition property of equality in the elimination method of solving a system of linear equations? B. What happens when you use the elimination method to solve a system of linear equations consisting of two parallel lines? C. How would you use the multiplication property of equality to solve a system of linear equations? D. What is the first step in solving a system of linear equations that contains fractions? EXERCISE SET 7.3 548 Chapter 7 Systems of Linear Equations SCAN TO ACCESS Vocabulary Review Choose the correct words to fill in the blanks below. substitute solution add variable solve multiplication property of equality Solving a System of Linear Equations by the Elimination Method Step 1: Decide which to eliminate. Step 2: Use the on each equation separately to make the coefficients of the variable that is to be eliminated opposites. Step 3: the respective left and right sides of the system together.
eBook - PDF- Mark D. Turner, Charles P. McKeague(Authors)
- 2017(Publication Date)
- XYZ Textbooks(Publisher)
1. Don’t Let Your Intuition Fool You As you become more experienced and more successful in mathematics, you will be able to trust your mathematical intuition. For now, though, it can get in the way of success. For example, if you ask a beginning algebra student to “subtract 3 from − 5” many will answer − 2 or 2. Both answers are incorrect, even though they may seem intuitively true. 2. Test Properties About Which You Are Unsure From time to time you will be in a situation in which you would like to apply a property or rule, but you are not sure if it is true. You can always test a property or statement by substituting numbers for variables. For instance, I always have students that rewrite ( x + 3) 2 as x 2 + 9, thinking that the two expressions are equivalent. The fact that the two expressions are not equivalent becomes obvious when we substitute 10 for x in each one. When x = 10, the expression ( x + 3) 2 is (10 + 3) 2 = 13 2 = 169 When x = 10, the expression x 2 + 9 = 10 2 + 9 = 100 + 9 = 109 It is not unusual, nor is it wrong, to try occasionally to apply a property that doesn’t exist. If you have any doubt about generalizations you are making, test them by replacing variables with numbers and simplifying. 267 4.1 Linear Equations Learning Objectives Learning Objectives 4.1 In this section, we will learn how to: 1. Determine if an ordered pair is a solution to a system of linear equations. 2. Solve a linear system by graphing. 3. Identify an inconsistent system. 4. Identify a system with dependent equations. Solving Linear Systems by Graphing Introduction Two linear equations considered at the same time make up what is called a system of linear equations .
eBook - PDFBeginning and Intermediate Algebra
A Guided Approach
- Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
Solve an application by setting up and solving a system of three linear equations in three variables. Unless otherwise noted, all content on this page is © Cengage Learning. Copyright 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 580 CHAPTER 8 Solving Systems of Linear Equations and Inequalities Unless otherwise noted, all content on this page is © Cengage Learning. We now extend the definition of a linear equation to include equations of the form ax 1 by 1 cz 5 d where a , b , c , and d are numerical values ( d is a constant). The solution of a system of three linear equations with three variables is an ordered triple of numbers if the equations are independent and the system consistent. For example, the solution of the system • 2 x 1 3 y 1 4 z 5 20 3 x 1 4 y 1 2 z 5 17 3 x 1 2 y 1 3 z 5 16 1 x , y , z 2 , which equals 1 1, 2, 3 2 is the ordered triple 1 1, 2, 3 2 . Each equation is satisfied if x 5 1 , y 5 2 , and z 5 3 . 2 x 1 3 y 1 4 z 5 20 3 x 1 4 y 1 2 z 5 17 3 x 1 2 y 1 3 z 5 16 2 1 1 2 1 3 1 2 2 1 4 1 3 2 0 20 3 1 1 2 1 4 1 2 2 1 2 1 3 2 0 17 3 1 1 2 1 2 1 2 2 1 3 1 3 2 0 16 2 1 6 1 12 0 20 3 1 8 1 6 0 17 3 1 4 1 9 0 16 20 5 20 17 5 17 16 5 16 The graph of an equation of the form ax 1 by 1 cz 5 d is a flat surface called a plane . A system of three linear equations in three variables is consistent or inconsis-tent, depending on how the three planes corresponding to the three equations intersect. Figure 8-15 illustrates some of the possibilities. Determine whether the equation x 1 2 y 1 3 z 5 6 is satisfied by the following values.
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