1. Algebra 1 Ch 7.3 – Linear Systems by Combinations

2. Objective Students will solve systems of linear equations by using the combination method

3. Before we begin… Thus far we have looked at solving systems of linear equations by graphing and the substitution method… In this lesson we will look at combining the equations to eliminate one of the variables, then solve the equation, and use the solution to determine the value of the other variable… Again, to be successful here you must be able to make decisions, be organized and work through all the steps…

4. Vocabulary Linear Combinations – A linear combination of two equations is an equation obtained by adding one of the equations (or multiples of one of the equations) to the other equation The result is an equation with 1 variable.

5. Process 1. Arrange the equations with like terms in columns 2. Multiply one or both of the equations by a number to obtain coefficients that are opposites for one of the variables 3. Add the equations from Step 2. Combining like terms will eliminate one variable. Solve for the remaining variable. 4. Substitute the value obtained in Step 3 into either of the original equations and solve for the other variable 5. Check the solution in each of the original equations

6. Comments A key to being successful with this method is to analyze the equations first. You do not always need to do each step. For example, sometimes the equations are already arranged in columns. Let’s look at an example and some of the thought processes involved…

7. Example #1 Use the combination method to solve the system of linear equations 4x + 3y = 16 2x – 3y = 8 Equation # 1 Equation # 2

8. Step 1 Arrange the equations with like terms in columns 4x + 3y = 16 2x – 3y = 8 Equation # 1 Equation # 2 In this example you do not have to do step 1 because the equations are already arranged with like terms in columns Remember – the sign in front of the term stays with the term! x’sy’s

9. Step 2 Multiply one or both of the equations by a number to obtain coefficients that are opposites for one of the variables 4x + 3y = 16 2x – 3y = 8 Equation # 1 Equation # 2 In this example you do not have to do step 2 because the equations have coefficients that are already opposites of each other (+3y and -3y)

10. Step 3 Add the equations from Step 2. Combining like terms will eliminate one variable. Solve for the remaining variable. 4x + 3y = 16 2x – 3y = 8 Equation # 1 Equation # 2 6x = 24 Add Equations Solve for the variable 6 x = 4 Solution

11. Step 4 Substitute the value obtained in Step 3 into either of the original equations and solve for the other variable x = 4 Value from Step 3 4x + 3y = 16 Equation # 1 Substitute the Value from Step 3 and solve for the other variable 4(4) + 3y = 16 16 + 3y = 16 -16 3y = 0 y = 0 Solution

12. Step 5 Check the solution in each of the original equations Solution x = 4y = 0 4x + 3y = 162x – 3y = 8 Equation # 1Equation # 2 4(4) + 3(0) = 16 16 + 0 = 16 16 = 16 True 2(4) – 3(0) = 8 8 – 0 = 8 8 = 8 True Since the ordered pair (4, 0) makes both statements true it is the solution to this system of equations

13. Example #2 Use combinations to solve the linear system of equations 3x + 5y = 6 Equation #1 -4x + 2y = 5 Equation #2

14. Step 1 Arrange the equations with like terms in columns 3x + 5y = 6 Equation #1 Equation #2 -4x + 2y = 5 In this example you do not have to do step 1 because the terms are already arranged in columns

15. Step 2 Multiply one or both of the equations by a number to obtain coefficients that are opposites for one of the variables 3x + 5y = 6 Equation #1 Equation #2 -4x + 2y = 5 In this example neither of the variables are opposite of each other. Therefore, you will need to multiply both equations by a number to get coefficients that are opposites… At this point you need to make some decisions…you can multiply to eliminate either the x’s or y’s you get to decide… I choose to multiply to get rid of the x’s…that means I will need to multiply equation #1 by 4 and equation #2 by 3 Be careful here…you will need to multiply the whole equation by what ever factor you choose

16. Step 2 (Continued) 3x + 5y = 6 Equation #1 Equation #2 -4x + 2y = 5 Multiply by 4 (3x + 5y = 6)(4) Revised Equation 12x + 20y =24 Multiply by 3 (3)(-4x + 2y = 5) Revised Equation -12x + 6y = 15

17. Step 3 Add the equations from Step 2. Combining like terms will eliminate one variable. Solve for the remaining variable. 12x + 20y =24 -12x + 6y = 15 Add and solve for y 26y = 39 26 y = 1.5

18. Step 4 Substitute the value obtained in Step 3 into either of the original equations and solve for the other variable Value from Step 3 y = 1.5 Substitute the value from step 3 and solve for the remaining variable -4x + 2y = 5 -4x + 2(1.5) = 5 -4x + 3 = 5 -3 -4x = 2 -4 x = -0.5Note: Be careful of the signs!...If the sign is wrong the whole problem is wrong!

19. Step 5 Check the solution in each of the original equations Solution Equation # 1Equation # 2 True Since the ordered pair (-0.5,1.5) makes both statements true it is the solution to this system of equations y = 1.5x = -0.5 3x + 5y = 6-4x + 2y = 5 3(-0.5) + 5(1.5) = 6 -1.5 + 7.5 = 6 6 = 6 -4(-0.5) + 2(1.5) = 5 2 + 3 = 5 5 = 5

20. Example # 3 Use linear combinations to solve the following system of equations 3x + 2y = 8Equation #1 2y = 12 – 5xEquation #2

21. Step 1 Arrange the equations with like terms in columns 3x + 2y = 8 Equation #1 2y = 12 – 5x Equation #2 In this example, after analyzing the equations you can see that the terms are not arranged in columns. You will need to transform equation #2 into standard form to get the terms in columns 2y = 12 – 5x +5x + 5x 5x + 2y = 12 Transformed Equation

22. Step 2 Multiply one or both of the equations by a number to obtain coefficients that are opposites for one of the variables 3x + 2y = 8 Equation #1 Equation #2 5x + 2y = 12 At this point, I see that I have +2y in both equations. You can multiply one of the equations by -1 to get a negative 2y It doesn’t matter which equation you multiply by -1. I choose to multiply equation #2 by -1 (5x + 2y = 12)(-1) -5x - 2y = -12 Revised Equation # 2 Don’t forget to multiply the WHOLE equation by -1!

23. Step 3 Add the equations from Step 2. Combining like terms will eliminate one variable. Solve for the remaining variable. 3x + 2y = 8 Equation #1 Equation #2 -5x - 2y = -12 Add the two equations together and then solve for the remaining variable -2x = - 4 -2 x = 2 Solution

24. Step 4 Substitute the value obtained in Step 3 into either of the original equations and solve for the other variable Value from Step 3 x = 2 3x + 2y = 8 Equation #1 Substitute the value from step 3 and solve for the remaining variable 3(2) + 2y = 8 6 + 2y = 8 - 6 -6 2y = 2 y = 1 Solution

25. Step 5 Check the solution in each of the original equations Solution x = 2 y = 1 3x + 2y = 8 Equation #1 2y = 12 – 5x Equation #2 3(2) + 2(1) = 8 6 + 2 = 8 8 = 8 True 2(1) = 12 – 5(2) 2 = 12 – 10 2 = 2 Since the ordered pair (2, 1) makes both statements true it is the solution to this system of equations

26. Comments On the next couple of slides are some practice problems…The answers are on the last slide… Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error… If you cannot find the error bring your work to me and I will help…

27. Your Turn Solve each system of equations by linear combinations 1. a – b = 8anda + b = 20 2. m + 3n = 2and -m +2n = 3 3. x – y = 0and -3x – y = 2 4. 2g – 3h = 0and 3g – 2h = 5 5. 10m + 16n = 140and 5m – 8n = 60

28. Your Turn 6. x + 3y = 12and -3y + x = 30 7. y = x – 9andx + 8y = 0 8. 0.1g + h + 4.3 = 0and 3.6 = -0.2g + h 9. 3a + 9b = 8b – aand5a – 10b = 4a -9b + 5 10. 9g – 7h = 2/3and3g + h = 1/3

29. Your Turn Solutions 1. (14, 6) 2. (-1,1) 3. (- ½, - ½ ) 4. (3, 2) 5. (13, 5/8) 6. (21, -3) 7. (8, -1) 8. (7, 5) 9. (1, -4) 10. (1/10, 1/30)

30. Summary A key tool in making learning effective is being able to summarize what you learned in a lesson in your own words… In this lesson we talked about solving linear systems using combinations. Therefore, in your own words summarize this lesson…be sure to include key concepts that the lesson covered as well as any points that are still not clear to you… I will give you credit for doing this lesson…please see the next slide…

31. Credit I will add 25 points as an assignment grade for you working on this lesson… To receive the full 25 points you must do the following: Have your name, date and period as well a lesson number as a heading. Do each of the your turn problems showing all work Have a 1 paragraph summary of the lesson in your own words Please be advised – I will not give any credit for work submitted: Without a complete heading Without showing work for the your turn problems Without a summary in your own words…