Chapter 7 Systems of Equations and Inequalities Copyright © 2014, 2010, 2007 Pearson Education, Inc Systems of Linear Equations in Two Variables

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2 Decide whether an ordered pair is a solution of a linear system. Solve linear systems by substitution. Solve linear systems by addition. Identify systems that do not have exactly one ordered- pair solution. Solve problems using systems of linear equations. Objectives:

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 3 Systems of Linear Equations and Their Solutions All equations in the form Ax + By = C are straight lines when graphed. Two such equations are called a system of linear equations or a linear system. A solution to a system of linear equations in two variables is an ordered pair that satisfies both equations in the system. A linear system that has at least one solution is called a consistent system. A linear system with no solution is called an inconsistent system.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4 Example: Determining Whether Ordered Pairs are Solutions of a System Consider the system: Determine if the ordered pair (1, 2) is a solution of the system. true The ordered pair (1, 2) satisfies both equations. (1, 2) is a solution of the system.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5 Example: Determining Whether Ordered Pairs Are Solutions of a System Consider the system: Determine if the ordered pair (7, 6) is a solution of the system. true false The ordered pair (7, 6) fails to satisfy both equations. Thus, the ordered pair is not a solution of the system.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6 Solving Linear Systems by Substitution

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7 Example: Solving a System by Substitution Solve by the substitution method: Step 1 Solve either of the equations for one variable in terms of the other. Step 2 Substitute the expression from step 1 into the other equation.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8 Example: Solving a System by Substitution (continued) Solve by the substitution method: Step 3 Solve the resulting equation containing one variable.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9 Example: Solving a System by Substitution (continued) Solve by the substitution method: Step 4 Back-substitute the obtained value into one of the original equations. The proposed solution is (–2, 5).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10 Example: Solving a System by Substitution (continued) Solve by the substitution method: Step 5 Check true The ordered pair (–2, 5) satisfies both equations. The solution set for this system of equations is {(–2, 5)}.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11 Solving Linear Systems by Addition

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 12 Example: Solving a System by the Addition Method Solve by the addition method: Step 1 Rewrite both equations in the form Ax + By = C. Both equations are already in this form. Variable terms appear on the left and constants appear on the right.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 13 Example: Solving a System by the Addition Method (continued) Solve by the addition method: Step 2 If necessary, multiply either equation or both equations by appropriate numbers so that the sum of the x-coefficients or the sum of the y-coefficients is 0.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 14 Example: Solving a System by the Addition Method (continued) Solve by the addition method: Step 3 Add the equations.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 15 Example: Solving a System by the Addition Method (continued) Solve by the addition method: Step 4 Solve the equation in one variable. Step 5 Back-substitute and find the value for the other variable. The proposed solution is (2, –1).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 16 Example: Solving a System by the Addition Method (continued) Solve by the addition method: Step 6 Check. true The solution set is {(2, –1)}.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 17 The Number of Solutions to a System of Two Linear Equations

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 18 Example: A System with No Solution Solve the system: The false statement 0 = 15 indicates that the system is inconsistent and has no solution. The solution set is the empty set,

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 19 Example: A System with No Solution (continued) Solve the system: We found that the solution for this system is the empty set, The lines are parallel and have no point of intersection.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 20 Example: A System with Infinitely Many Solutions Solve the system: In our final step, both variables have been eliminated and the resulting statement, –40 = –40, is true. This true statement indicates that the system has infinitely many solutions.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 21 Example: A System with Infinitely Many Solutions (continued) Solve the system: We found that there are infinitely many solutions. The solution set for this system may be expressed as: or: The equations represent the same line.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 22 Example: Finding a Break-Even Point A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. a. Write the cost function, C, of producing x pairs of running shoes. b. Write the revenue function, R, from the sale of x pairs of running shoes.

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 23 Example: Finding a Break-Even Point (continued) A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. Determine the break-even point. Describe what this means. The break-even point occurs where the graphs of C and R intersect. Thus, we find this point by solving the system: or

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 24 Example: Finding a Break-Even Point A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. Determine the break-even point. The break-even point is (6000, 480,000).

Copyright © 2014, 2010, 2007 Pearson Education, Inc. 25 Example: Finding a Break-Even Point (continued) A company that manufactures running shoes has a fixed cost of $300,000. Additionally, it costs $30 to produce each pair of shoes. They are sold at $80 per pair. c. The break-even point is (6000,480,000). Describe what this means. This means that the company will break even if it produces and sells 6000 pairs of running shoes. At this level, the money coming in is equal to the money going out: $480,000.