Mathematics
Simultaneous Equations
Simultaneous equations are a set of equations with multiple variables that are solved together to find the values of the variables that satisfy all the equations simultaneously. This is typically done using methods like substitution, elimination, or matrices. Solving simultaneous equations allows for finding the intersection point of two lines or the common solution to multiple equations.
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10 Key excerpts on "Simultaneous Equations"
eBook - PDF- William R. Gondin, Bernard Sohmer(Authors)
- 2014(Publication Date)
- Made Simple(Publisher)
These are discussed in Chapters VI and VII. This present chapter is mainly concerned with linear equations in two unknowns, illustrated by the second example above. A set of equations which are to be considered together in connexion with the same problem is called a system of Simultaneous Equations; or, for short, a system of equations. A solution of an equation, or of a system of equations, is a set of values for the unknown quantities which satisfy the mathematical conditions expressed by the equation or equations. When only one such set of values for the unknowns is possible the solution is said to be a unique solution. When all possible solutions for a system of equations have been found the system is said to have been solved simultaneously. Examples for systems of simultaneous linear equations in two unknowns follow throughout the rest of this chapter. Examples for other types of equations follow in later chapters. 15 16 Intermediate Algebra and Analytic Geometry Made Simple Systems of Two Equations From elementary algebra you should already know how a system of two linear equations in two unknowns may be solved simultaneously. To review: EXAMPLE 1 : Solve simultaneously, x -2y = -8 (1) x + y = 7 (2) SOLUTION (by the method of comparison) : -3>> = -15 7 = 5 x + 5 = 7 x = 2 Answer: x — 2, y = (Subtracting equation (2) from equation (1) by axiom 2, page 10) (Dividing by — 3 = —3, axiom 4) (Substituting y = 5 in equation (2)) (Transposing 5; that is: subtracting 5 = 5 from the above equation by axiom 2) = 5 SOLUTION (by the method of substitution): x = 2y - 8 8 + ^ = 7 3y = 15 7 = 5 x = 2(5) - 8 x = 2 (Transposing 2y in equation (1)) (Substitution for x in equation (2)) (Transposing, etc.) (Dividing by 3 = 3) (Substituting y = 5 in x = 2y — 8) (Removing parentheses, etc.) Answer: x — 2, y = 5 How do we know that the result arrived at by the above methods is really a solution of the system? The arithmetic method of verifying your work is check by substitution.
Available until 5 Dec |Learn more- Mike Rosser, Piotr Lis(Authors)
- 2016(Publication Date)
- Routledge(Publisher)
5 Simultaneous linear equations DOI: 10.4324/9781315641713-5Learning objectives
After completing this chapter students should be able to:- Solve sets of simultaneous linear equations with two or more variables using the substitution and row operations methods.
- Relate simultaneous linear equations mathematical solutions to economic analysis, including supply and demand and the basic Keynesian macroeconomic models.
- Construct and use break-even charts.
- Recognize when a linear equations system cannot be solved.
- Derive the reduced form equations for the equilibrium values of dependent variables in basic linear economic models and interpret their meaning.
- Derive the profit maximizing solutions to price discrimination and multiplant monopoly problems involving linear functions.
- Set up linear programming constrained maximization and minimization problems and solve them using the graphical method.
5.1 Systems of Simultaneous Linear Equations
The way to solve single linear equations with one unknown was explained in Chapter 3 . We now turn to sets of linear equations with more than one unknown. A simultaneous linear equation system exists when:- there is more than one functional relationship between a set of specified variables, and
- all the functional relationships are in a linear form.
1the demand schedule isp = 420 − 0.2 q2and the supply schedule isp = 60 + 0.4 qIf this market is in equilibrium then the equilibrium price and quantity will be where the demand and supply schedules intersect. As this will correspond to a point which is on both the demand schedule and the supply schedule then the equilibrium values of p and q
eBook - PDFEssential Maths
for Business and Management
- Clare Morris(Author)
- 2007(Publication Date)
- Bloomsbury Academic(Publisher)
By plotting graphs of the various charging regimes and finding where the graphs intersect, you could work out what volume of calls would make it cheaper for you to switch to a different provider. More generally, it’s clear that in the complex world of organisations and management, very few problems will be able to be represented by just a single equation with one unknown quantity. It’s much more likely that sets of equations, representing groups of interacting factors, will be required to give a true model of reality. So simultaneous equa-tions, and more advanced topics relating to sets of equations, have an important part to play in the modelling of business problems. Solving simultaneous linear equations If you have come across the idea of Simultaneous Equations before, then it was probably in relation to pairs of simultaneous linear equations, like the following: x + 2 y = 5 3 x + 5 y = 11 These are linear equations, because they contain no powers of x or y higher than the first, no xy terms, and so on. However, there are now two unknown variables – x and y – whereas up to now we’ve only dealt with equations with one variable. Will we be able to solve for both x and y ? The answer is yes – because we have not one, but two equations linking our two vari-ables. Just as we could solve one equation with one unknown variable, so in general we need two equations to give us a numerical solution for two variables (and so on, although we won’t be dealing with more than two). There are two methods for solving pairs of equations of this kind, and we are going to cover both of them, because you might have come across either if you’ve met this topic before. We’ll illustrate both with the pair of equations above. (a) The elimination method This method gets its name because we eliminate one variable from the equations and solve for the one that’s left.
- Yogesh Jaluria(Author)
- 2011(Publication Date)
- CRC Press(Publisher)
171 6 Numerical Solution of Simultaneous Algebraic Equations 6.1 INTRODUCTION Systems of simultaneous algebraic equations are frequently encountered in engineering applications such as those concerned with electrical networks, structural analysis, heat transfer, fluid flow, optimization, vibrations, chemical reactions, and data analysis. The numerical solution of an ODE or a PDE also often reduces to the solution of a set of algebraic equations, as discussed later in Chapters 9 and 10. A system of n Simultaneous Equations, with x 1 , x 2 , . . ., x n as the n unknowns, may be written as f x x x f x x x f x x x n n n n 1 1 2 2 1 2 1 2 0 0 0 ( , , , ) ( , , , ) ( , , , ) … … vertellipsis vertellipsis … = = = (6.1) where f 1 , f 2 , . . ., f n denote n different functions of the n independent variables. Various methods have been developed to solve this system of equations to obtain the values of the variables x 1 , x 2 , . . ., x n . The choice of a particular method for a given problem generally depends on the nature of the equations and the number of unknowns n. In many circumstances, the equations are linear in the unknown variables. Such a system of linear equations has the general form a x a x a x b a x a x a x b a x a x n n n n n n 11 1 12 2 1 1 21 1 22 2 2 2 1 1 2 2 + + + = + + + = + + midhorizellipsis midhorizellipsis vertellipsis vertellipsis midhorizellipsis + = a x b nn n n (6.2) where the a ’s represent n 2 coefficients and the b ’s similarly represent n constants. In matrix notation, this system may be written more concisely as AX B = (6.3) 172 Computer Methods for Engineering with MATLAB ® Applications where A is a square matrix of the coefficients, X is a column matrix, or vector, of the unknowns, and B is a column matrix, or vector, of the constants that appear on the right-hand side of the equations. From Equation 6.2, a ij represents an element of the matrix A and b i an element of the vector B.
eBook - PDF- Marvin Tobias(Author)
- 2022(Publication Date)
- Springer(Publisher)
83 C H A P T E R 4 Linear Simultaneous Equation Sets 4.1 INTRODUCTION This chapter turns to an interpretation of the solution to linear equation sets, using a geometric approach and insight. We will look at an equation set in several different (and perhaps new) ways, and consider the solvability and compatibility of an equation set. Most of the mechanics of solution have already been discussed. This chapter intends to be largely conceptual. Many applications in mechanics, dynamics, and electric circuits depend on the insights gained, and presented here. We begin by defining the equation set Ax = b as “nonhomogeneous” because the b vector is assumed to be nonzero. Associated with this set is the “homogeneous” set, Ax = 0; the same set, but with the b vector replaced by the zero vector. In the event that matrix A is nonsingular, and has an inverse, the homogeneous set plays no part. But, when A is singular, we will find interest in both Ax = 0, and in A x = 0 (the transposed homogeneous set). 4.2 VECTORS AND VECTOR SETS In order to gain greater insight into its solution, the equation set will be interpreted as a “vector transformation.” The equation Ax = y “transforms” the columns of A(nXm) into the vector y. Alternatively, y is “synthesized” as a linear vector sum of the column vectors of A. Ax = y = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ a 11 a 21 . . . a n1 ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ x 1 + ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ a 12 a 22 . . . a n2 ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ x 2 + · · · + ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ a 1m a 2m . . . a nm ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ x m = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ y 1 y 2 . . . y n ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭ , or: (4.1) = a 1 x 1 + a 2 x 2 + · · · + a m x m = y . (4.2) In this quite general example, A is (nXm); there are m vectors (the columns of A), each with n coordinates (dimensions — the rows of A). It is often instructive to draw the same vector picture of the transposed matrix, i.e., A , whose n column vectors are the rows of A.
eBook - PDF- Jerome Kaufmann, Karen Schwitters, , , Jerome Kaufmann, Karen Schwitters(Authors)
- 2014(Publication Date)
- Cengage Learning EMEA(Publisher)
We illustrate these cases in Figure 10.2. y (4, 2) x x + y = 6 x - y = 2 Figure 10.1 Chapter Preview Equations in two variables were previously introduced in Chapter 7. This chapter considers systems of two or more equations. That is, we are looking for solutions that satisfy more than one equation at the same time. Systems of equations are also referred to as Simultaneous Equations. Three methods for solving systems covered in this chapter are • Graphing • Substitution method • Elimination-by-addition method Copyright 2013 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. 10.1 • Systems of Two Linear Equations and Linear Inequalities in Two Variables 505 Unless otherwise noted, all content on this page is © Cengage Learning. Case I one solution Case II no solutions Case III infinitely many solutions x y y y x x Figure 10.2 Case I The graphs of the two equations are two lines intersecting in one point, and there is one solution . This system is called a consistent system . Case II The graphs of the two equations are parallel lines, and there is no solution . This system is called an inconsistent system . Case III The graphs of the two equations are the same line, and there are infinitely many solutions to the system. Any pair of real numbers that satisfies one of the equa-tions will also satisfy the other equation. We say that the equations are dependent . Thus as we solve a system of two linear equations in two variables, we know what to expect. The system will have no solutions, one ordered pair as a solution, or infinitely many ordered pairs as solutions.
eBook - PDF- Mark D. Turner, Charles P. McKeague(Authors)
- 2016(Publication Date)
- XYZ Textbooks(Publisher)
269 4.1 Learning Objectives In this section we will learn how to: 1. Solve a system of linear equations in two variables graphically. 2. Solve a system of linear equations in two variables using the addition method. 3. Solve a system of linear equations in two variables using substitution. 4. Recognize an inconsistent system or a system with dependent equations. Systems of Linear Equations in Two Variables Introduction Previously, we found the graph of an equation of the form ax + by = c to be a line. Because the graph is a line, the equation is said to be a linear equation. Two linear equations considered together form a linear system of equations. For example, 3 x − 2 y = 6 2 x + 4 y = 20 is a linear system. The solution set to the system is the set of all ordered pairs that satisfy both equations. If we graph each equation on the same set of axes, we can see the solution set (see Figure 1). The point (4, 3) lies on both lines and therefore must satisfy both equations. It is obvious from the graph that it is the only point that does so. The solution set for the system is {(4, 3)}. More generally, if a 1 x + b 1 y = c 1 and a 2 x + b 2 y = c 2 are linear equations, then the solution set for the system a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2 can be illustrated through one of the graphs in Figure 2. FIGURE 1 3 x 2 y 6 3 x -2 y = 6 2 x 4 y 20 2 x + 4 y = 20 x y 1 -1 -2 -3 -4 -5 3 4 5 2 1 -3 -5 2 3 4 5 -2 -1 (4, 3) (4, 3) 270 CHAPTER 4 Systems of Equations Case I The two lines intersect at one and only one point. The coordinates of the point give the solution to the system. This is what usually happens. Case II The lines are parallel and therefore have no points in common. The solution set to the system is the empty set, ∅ . In this case, we say the equations are inconsistent . Case III The lines coincide. That is, their graphs represent the same line. The solution set consists of all ordered pairs that satisfy either equation.
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.
eBook - PDF- Mark D. Turner, Charles P. McKeague(Authors)
- 2017(Publication Date)
- XYZ Textbooks(Publisher)
two lines that intersect at a single point d. two lines with one of the lines vertical 46. If the graph of a system of linear equations consists of two lines that coincide, then the solution set: a. is all real numbers b. is the empty set c. contains an infinite number of ordered pairs d. is undefined Getting Ready For the Next Section Simplify each of the following. 47. ( x + y ) + ( x − y ) 48. ( x + 2 y ) + ( − x + y ) 49. 3(2 x − y ) + ( x + 3 y ) 50. 3(2 x + 4 y ) − 2(3 x + 5 y ) 51. − 4(3 x + 5 y ) + 5(5 x + 4 y ) 52. (3 x + 8 y ) − (3 x − 2 y ) 53. 6 1 __ 2 x − 1 __ 3 y 54. 12 1 __ 4 x + 2 __ 3 y 55. Let x + y = 4. If x = 3, find y . 56. Let x + 2 y = 4. If y = 3, find x . 57. Let x + 3 y = 3. If x = 3, find y . 58. Let 2 x + 4 y = − 1. If y = 1 _ 2 , find x . 59. Let 3 x + 5 y = − 7. If x = 6, find y . 60. Let 3 x − 2 y = 12. If y = 6, find x . x y 1 –1 –3 –4 –5 3 4 5 3 2 –2 1 –3 –4 –5 2 4 5 –2 –1 Solution Solution x y 1 –1 –3 –4 –5 3 4 5 3 2 –2 1 –3 –4 –5 2 4 5 –2 –1 Solution Solution x y 1 –1 –3 –4 –5 3 4 5 3 2 –2 1 –3 –4 –5 2 4 5 –2 –1 Solution Solution x y 1 –1 –3 –4 –5 3 4 5 3 2 –2 1 –3 –4 –5 2 4 5 –2 –1 Solution Solution 277 4.2 Linear Equations Learning Objectives Learning Objectives 4.2 In this section, we will learn how to: 1. Solve a system of linear equations using the elimination method. The Elimination Method Introduction The addition property of equality states that if equal quantities are added to both sides of an equation, the solution set is unchanged. In the past we have used this property to help solve equations in one variable. We will now use it to solve systems of linear equations. Here is another way to state the addition property of equality. Let A , B , C , and D represent algebraic expressions. If A = B and C = D then A + C = B + D Because C and D are equal (that is, they represent the same number), what we have done is added the same amount to both sides of the equation A = B .
eBook - PDFIntroductory Algebra
Concepts with Applications
- Charles P. McKeague(Author)
- 2013(Publication Date)
- XYZ Textbooks(Publisher)
Step 2: Graph the second equation on the same set of axes. Step 3: Read the coordinates of the point where the graphs cross each other (the coordinates of the point of intersection.) Step 4: Check the solution to see that it satisfies both equations. Strategy for Solving a System by the Substitution Method [7.2] Step 1: Solve either of the equations for one of the variables (this step is not necessary if one of the equations has the correct form already.) Step 2: Substitute the results of step 1 into the other equation, and solve. Step 3: Substitute the results of step 2 into an equation with both x and y variables, and solve. (The equation produced in step 1 is usually a good one to use.) Step 4: Check your solution, if necessary. EXAMPLES 1. The solution to the system x + 2y = 4 x − y = 1 is the ordered pair (2, 1). It is the only ordered pair that satis- fies both equations. 2. Solving the system in Example 1 by graphing looks like x y (2, 1) x + 2y = 4 x − y = 1 3. We can apply the substitution method to the system in Exam- ple 1 by first solving the second equation for x to get x = y + 1. Substituting this expression for x into the first equation, we have (y + 1) + 2y = 4 3y + 1 = 4 3y = 3 y = 1 Using y = 1 in either of the original equations gives x = 2. 565 Chapter 7 Summary Strategy for Solving a System by the Elimination Method [7.3] Step 1: Look the system over to decide which variable will be easier to eliminate. Step 2: Use the multiplication property of equality on each equation separately to ensure that the coefficients of the variable to be eliminated are opposites. Step 3: Add the left and right sides of the system produced in step 2, and solve the resulting equation. Step 4: Substitute the solution from step 3 back into any equation with both x and y variables, and solve. Step 5: Check your solution in both equations, if necessary. Special Cases [7.1, 7.2, 7.3] In some cases, using the elimination or substitution method eliminates both variables.
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