1. Copyright © 2011 Pearson Education, Inc. Slide 7.1-1

2. Copyright © 2011 Pearson Education, Inc. Slide 7.1-2 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 © 2011 Pearson Education, Inc. Slide 7.1-3 7.1 Systems of Equations A linear equation in n unknowns has the form where the variables are of first-degree. A set of equations is called a system of equations. The solutions of a system of equations must satisfy every equation in the system. If all equations in a system are linear, the system is a system of linear equations, or a linear system.

4. Copyright © 2011 Pearson Education, Inc. Slide 7.1-4 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 © 2011 Pearson Education, Inc. Slide 7.1-5 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 © 2011 Pearson Education, Inc. Slide 7.1-6 7.1 Solving a Linear System in Two Variables Graphically ExampleSolve the system graphically. SolutionSolve (1) and (2) for y. (1) (2)

7. Copyright © 2011 Pearson Education, Inc. Slide 7.1-7 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 © 2011 Pearson Education, Inc. Slide 7.1-8 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 © 2011 Pearson Education, Inc. Slide 7.1-9 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 © 2011 Pearson Education, Inc. Slide 7.1-10 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 © 2011 Pearson Education, Inc. Slide 7.1-11 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 © 2011 Pearson Education, Inc. Slide 7.1-12 7.1 Solving a Nonlinear System of Equations Substitute these values for x into (3). The solution set is

13. Copyright © 2011 Pearson Education, Inc. Slide 7.1-13 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 © 2011 Pearson Education, Inc. Slide 7.1-14 7.1 Applications of Systems Solving Application Problems 1.Read the problem carefully and assign a variable to represent each quantity. 2.Write a system of equations involving these variables. 3.Solve the system of equations and determine the solution. 4.Look back and check your solution. Does it seem reasonable?

15. Copyright © 2011 Pearson Education, Inc. Slide 7.1-15 7.1 Applications of Systems 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.

16. Copyright © 2011 Pearson Education, Inc. Slide 7.1-16 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.