1. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Section 7.1 Systems of Linear Equations

2. Copyright 2013, 2010, 2007, Pearson, Education, Inc. What You Will Learn Systems of Linear Equations Solutions to Systems of Linear Equations Solving Systems of Linear Equations by Graphing 7.1-2

3. Copyright 2013, 2010, 2007, Pearson, Education, Inc. System of Linear Equations When two or more linear equations are considered simultaneously, the equations are called a system of linear equations. A solution to a system of equations is the ordered pair or pairs that satisfy all equations in the system. A system of linear equations may have exactly one solution, no solution, or infinitely many solutions. 7.1-3

4. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Is the Ordered Pair a Solution? Determine which of the ordered pairs is a solution to the following system of linear equations. 2x – y = 6 2x + y = 14 a. (0, –6)b. (5, 4)c. (3, 8) 7.1-4

5. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Is the Ordered Pair a Solution? Solution Check (0, –6) (0,–6) does not satisfy both equations, it is not a solution to the system. TrueFalse 7.1-5

6. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Is the Ordered Pair a Solution? Solution Check (5, 4) (5, 4) satisfies both equations, it is a solution to the system. True 7.1-6

7. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 1: Is the Ordered Pair a Solution? Solution Check (3, 8) (3, 8) does not satisfy both equations, it is not a solution to the system. FalseTrue 7.1-7

8. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Solving a System of Equations by Graphing 1.Determine three ordered pairs that satisfy each equation. 2.Plot the ordered pairs and sketch the graphs of both equations on the same axes. 3.The coordinates of the point or points of intersection of the graphs are the solution or solutions to the system of equations. 7.1-8

9. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations When graphing linear equations, three outcomes are possible: 1.The two lines may intersect at one point, producing a system with one solution. A system that has one solution is called a consistent system of equations. 7.1-9

10. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations 7.1-10

11. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations When graphing linear equations, three outcomes are possible: 2.The two lines may be parallel, producing a system with no solutions. A system with no solutions is called an inconsistent system. 7.1-11

12. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations 7.1-12

13. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations When graphing linear equations, three outcomes are possible: 3.The two equations may represent the same line, producing a system with an infinite number of solutions. A system with an infinite number of solutions is called a dependent system. 7.1-13

14. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Graphing Linear Equations 7.1-14

15. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: A System with One Solution Determine the solution to the following system of equations graphically. 7.1-15

16. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: A System with One Solution Solution Ordered pairs that satisfy x + y = 4 are: (0, 4), (1, 3), (4, 0) Ordered pairs that satisfy 2x – y = –1 are: (–2, –3), (0, 1), (1, 3) Graph both lines on the same axes 7.1-16

17. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: A System with One Solution Solution 7.1-17

18. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 2: A System with One Solution Solution The graphs intersect at (1, 3), which is the solution to the system of equations. This point is the only point that satisfies both equations. True 7.1-18

19. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: A System with No Solution Determine the solution to the following system of equations graphically. 7.1-19

20. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: A System with No Solution Solution Ordered pairs that satisfy x + y = 3 are: (0, 3), (3, 0), (1, 2) Ordered pairs that satisfy 2x + 2y =–4 are: (–2, 0), (0, –2), (1, –3) Graph both lines on the same axes 7.1-20

21. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: A System with No Solution Solution 7.1-21

22. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 3: A System with No Solution Solution Since the two lines are parallel, they do not intersect; therefore, the system has no solution. 7.1-22

23. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business At a collectibles show, Richard Lane can sell model trains for $35. The costs for making the trains are a fixed cost of $200 and a production cost of $15 apiece. a)Write an equation that represents Richard’s revenue. Write an equation that represents Richard’s cost. 7.1-23

24. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business b)How many model trains must Richard sell to break even? c)Write an equation for the profit formula. Use the formula to determine Richard’s profit if he sells 15 model trains. d)How many model trains must Richard sell to make a profit of $600? 7.1-24

25. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business Solution a)Let x denote the number of model trains made and sold. Revenue: R = 35x ($35 times # of trains) Cost: C = 200 + 15x (200 plus $15 times # of trains) 7.1-25

26. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business Solution b)Break-even point is where graphs intersect (10, 350) He must make and sell 10 model trains. Cost = $350 Revenue = $350 7.1-26

27. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business Solution c)Profit = Revenue – Cost For 15 trains, profit is: 7.1-27

28. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business Solution c)Richard has a profit of $100 if he sells 15 trains. Observe the graph. At 15 trains the revenue line is greater than the cost line, that is, there is a profit. 7.1-28

29. Copyright 2013, 2010, 2007, Pearson, Education, Inc. Example 6: Modeling–Profit and Loss in Business Solution d)Substitute $600 for P, solve for x Richard must sell 40 model trains to make a profit of $600. 7.1-29