Copyright © 2007 Pearson Education, Inc. Slide 7-2 Chapter 7: Systems of Equations and Inequalities; Matrices 7.1Systems of Equations 7.2Solution of Linear.

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2 Copyright © 2007 Pearson Education, Inc. Slide 7-2 Chapter 7: Systems of Equations and Inequalities; Matrices 7.1Systems of Equations 7.2Solution of Linear Systems in Three Variables 7.3Solution of Linear Systems by Row Transformations 7.4Matrix Properties and Operations 7.5Determinants and Cramer’s Rule 7.6Solution of Linear Systems by Matrix Inverses 7.7Systems of Inequalities and Linear Programming 7.8Partial Fractions

3 Copyright © 2007 Pearson Education, Inc. Slide 7-3 7.1 Systems of Equations A set of equations is called a system of equations. The solutions must satisfy each equation in the system. A linear equation in n unknowns has the form where the variables are of first-degree. If all equations in a system are linear, the system is a system of linear equations, or a linear system.

4 Copyright © 2007 Pearson Education, Inc. Slide 7-4 Three possible solutions to a linear system in two variables: 1.One solution: coordinates of a point, 2.No solutions: inconsistent case, 3.Infinitely many solutions: dependent case. 7.1 Linear System in Two Variables

5 Copyright © 2007 Pearson Education, Inc. Slide 7-5 7.1 Substitution Method ExampleSolve the system. Solution Solve (2) for y. Substitute y = x + 3 in (1). Solve for x. Substitute x = 1 in y = x + 3. Solution set: {(1, 4)} (1) (2)

6 Copyright © 2007 Pearson Education, Inc. Slide 7-6 7.1 Solving a Linear System in Two Variables Graphically ExampleSolve the system graphically. SolutionSolve (1) and (2) for y. (1) (2)

7 Copyright © 2007 Pearson Education, Inc. Slide 7-7 7.1 Elimination Method ExampleSolve the system. SolutionTo eliminate x, multiply (1) by –2 and (2) by 3 and add the resulting equations. (3) (4) (1) (2)

8 Copyright © 2007 Pearson Education, Inc. Slide 7-8 7.1 Elimination Method Substitute 2 for y in (1) or (2). The solution set is {(3, 2)}. Check the solution set by substituting 3 in for x and 2 in for y in both of the original equations.

9 Copyright © 2007 Pearson Education, Inc. Slide 7-9 7.1 Solving an Inconsistent System ExampleSolve the system. SolutionEliminate x by multiplying (1) by 2 and adding the result to (2). Solution set is . (1) (2) Inconsistent System

10 Copyright © 2007 Pearson Education, Inc. Slide 7-10 7.1 Solving a System with Dependent Equations ExampleSolve the system. SolutionEliminate x by multiplying (1) by 2 and adding the result to (2). Each equation is a solution of the other. Choose either equation and solve for x. The solution set is e.g. y = –2: (1) (2)

11 Copyright © 2007 Pearson Education, Inc. Slide 7-11 7.1 Solving a Nonlinear System of Equations ExampleSolve the system. SolutionChoose the simpler equation, (2), and solve for y since x is squared in (1). Substitute for y into (1). (1) (2)

12 Copyright © 2007 Pearson Education, Inc. Slide 7-12 7.1 Solving a Nonlinear System of Equations Substitute these values for x into (3). The solution set is

13 Copyright © 2007 Pearson Education, Inc. Slide 7-13 7.1 Solving a Nonlinear System Graphically ExampleSolve the system. Solution(1) yields Y 1 = 2 x ; (2) yields Y 2 = |x + 2|. The solution set is {(2, 4), (–2.22,.22), (–1.69,.31)}.

14 Copyright © 2007 Pearson Education, Inc. Slide 7-14 7.1 Applications of Systems To solve problems using a system 1.Determine the unknown quantities 2.Let different variables represent those quantities 3.Write a system of equations – one for each variable ExampleIn a recent year, the national average spent on two varsity athletes, one female and one male, was $6050 for Division I-A schools. However, average expenditures for a male athlete exceeded those for a female athlete by $3900. Determine how much was spent per varsity athlete for each gender.

15 Copyright © 2007 Pearson Education, Inc. Slide 7-15 7.1 Applications of Systems Solution Let x = average expenditures per male y = average expenditures per female Average spent on one male and one female Average Expenditure per male: $8000, and per female: from (2) y = 8000 – 3900 = $4100.