352 CHAPTER 6 Algebra, Graphs, and Functions Delbert is choosing between two membership options at the local movie society. A director membership has a lower initial cost to join, but the cost per movie is higher. A producer membership has a higher initial cost to join, but the cost per movie is lower. Which membership would be a better financial choice for Delbert? In this section, we will learn how to use two equations, considered simultaneously, to help answer this question. Solving Systems of Linear Equations SECTION 6.7 LEARNING GOALS Upon completion of this section, you will be able to: 7 Determine if an ordered pair is a solution to a system of linear equations. 7 Solve a system of linear equations by graphing. 7 Solve a system of linear equations by the substitution method. 7 Solve a system of linear equations by the addition method. 7 Solve applications of systems of linear equations. Why This Is Important As we will see in this section, we will use a system of equations to model many real-life problems including business profit or loss, medicine mixtures, and a variety of financial applications. These applications have relevance to a wide variety of occupations and to many aspects of our daily lives. In algebra, it is often necessary to determine the common solution to two or more linear equations. When two or more linear equations are considered simultaneously, the equations are called a system of linear equations. The solution to a system of linear equations may be determined by a number of techniques. In this section, we illustrate three different methods for solving systems of linear equations. Solutions to Systems of Linear Equations A solution to a system of equations is the ordered pair or ordered pairs that satisfy all equations in the system. A system of linear equations may have exactly one solution, no solution, or infinitely many solutions. Example 1 Is the Ordered Pair a Solution? Determine which of the ordered pairs is a solution to the system of equations. y x 1 2 4 = − x y 3 5 9 + = − a) (0, 4) − b) (12, 9) − c) (2, 3) − Solution To be a solution, the ordered pair must satisfy each equation in the system. a) For − (0, 4) we substitute x 0 = and y 4 = − into each equation in the system. y x 1 2 4 4 1 2 (0) 4 4 0 4 4 4 = − − = − − = − − = − x y 3 5 9 3(0) 5( 4) 9 0 20 9 20 9 + = − + − = − − = − − = − True False Since − (0, 4) does not satisfy both equations, it is not a solution to the system. Goncharov_Artem/Shutterstock
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