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2. Lesson Menu Five-Minute Check (over Lesson 6–2) CCSS Then/Now New Vocabulary Key Concept: Solving by Elimination Example 1:Elimination Using Addition Example 2:Write and Solve a System of Equations Example 3:Elimination Using Subtraction Example 4:Real-World Example: Write and Solve a System of Equations

3. CCSS Content Standards A.CED.2 Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales. A.REI.6 Solve systems of linear equations exactly and approximately (e.g., with graphs), focusing on pairs of linear equations in two variables. Mathematical Practices 7 Look for and make use of structure. Common Core State Standards © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

4. Then/Now You solved systems of equations by using substitution. Solve systems of equations by using elimination with addition. Solve systems of equations by using elimination with subtraction.

5. Concept

6. Example 1 Elimination Using Addition Use elimination to solve the system of equations. –3x + 4y = 12 3x – 6y = 18 Since the coefficients of the x-terms, –3 and 3, are additive inverses, you can eliminate the x-terms by adding the equations. Write the equations in column form and add. The x variable is eliminated. Divide each side by –2. y =–15Simplify.

7. Example 1 Elimination Using Addition Now substitute –15 for y in either equation to find the value of x. –3x + 4y=12First equation –3x + 4(–15)=12Replace y with –15. –3x – 60=12Simplify. –3x – 60 + 60=12 + 60Add 60 to each side. –3x=72Simplify. Divide each side by –3. x =–24Simplify. Answer: The solution is (–24, –15).

8. 4x + 6y = 32 3x – 6y = 3 Notice that you have a + 6y and a – 6y, they are the same variables with opposite coefficients Therefore, we can add the two equations together and eliminate the y variable 4x + 6y = 32 + 3x – 6y = 3 7x = 35 7 7 x = 5

9. Now that we have found the value of x is 5 we can substitute 5 in the place of x in either equation I use the 1 st equation 4x + 6y = 32 4(5) + 6y = 32 20 + 6y = 32 -20 -20 6y = 12 6 6 y = 2 (5, 2)

10. Example 1 Use elimination to solve the system of equations. 3x – 5y = 1 2x + 5y = 9 A.(1, 2) B.(2, 1) C.(0, 0) D.(2, 2)

11. Use elimination to solve the system of equations 1)-4x + 3y = -3 4x – 5y = 5 (0, -1) 2) 4y + 3x = 22 3x – 4y = 14 (6, 1)

12. 3) –v + w = 7 v + w = 1 (-3, 4) 4) -4x + 5y = 17 4x + 6y = -6 (-3, 1) 5) -3x – 8y = -24 3x – 5y = 4.5 ( 4, 1.5)

13. We can use elimination to find specific numbers that are described as being related to each other.

16. Example 2 Write and Solve a System of Equations Four times one number minus three times another number is 12. Two times the first number added to three times the second number is 6. Find the numbers. Let x represent the first number and y represent the second number. Four times one number minus three times another number is12. Two times the first number added to three times the second number is6. 4x4x – 3y3y = 12 2x2x + 3y3y = 6

17. Example 2 Write and Solve a System of Equations Use elimination to solve the system. x =3Simplify. Write the equations in column form and add. 6x = 18The y variable is eliminated. Divide each side by 6. 4x – 3y=12 (+) 2x + 3y= 6 Now substitute 3 for x in either equation to find the value of y.

18. Example 2 Write and Solve a System of Equations 4x – 3y=12First equation y=0Simplify. 4(3) – 3y =12Replace x with 3. 12 – 3y=12Simplify. 12 – 3y – 12 =12 – 12Subtract 12 from each side. –3y=0Simplify. Divide each side by –3. Answer: The numbers are 3 and 0.

19. Example 2 A.–3, 2 B.–5, –5 C.–5, –6 D.1, 1 Four times one number added to another number is –10. Three times the first number minus the second number is –11. Find the numbers.

20. 1)The sum of two numbers is -10. Negative three times the first number minus the second number equals 2. Find the number. 4, -14 2) The sum of two numbers is 22, and their difference is 12. What are the numbers? 5, 17 3) Find the two numbers with a sum of 41 and a difference of 9. 25, 16

21. Sometimes we can eliminate a variable by subtracting one equation from another. Solve the system of equations: 2t + 5r = 6 9r + 2t = 22 Put like variables underneath each other Since both equations contain 2t, we will use elimination by subtraction because they are not opposites

22. 2t + 5r = 6 (-) 2t + 9r = 22 subtracting this equation is the same as multiplying each term by -1 2t + 5r = 6 -2t – 9r = -22 -4r = -16 -4 -4 r = 4 Now substitute r with 4 in the first equation

23. 2t + 5r = 6 2t + 5(4) = 6 2t + 20 = 6 -20 -20 2t = -14 2 2 t = -7 (4, -7)

24. Example 3 Elimination Using Subtraction Use elimination to solve the system of equations. 4x + 2y = 28 4x – 3y = 18 Since the coefficients of the x-terms are the same, you can eliminate the x-terms by subtracting the equations. y=2Simplify. Write the equations in column form and subtract. 5y =10The x variable is eliminated. Divide each side by 5. 4x + 2y=28 (–) 4x – 3y=18

25. Example 3 Elimination Using Subtraction Now substitute 2 for y in either equation to find the value of x. Answer: The solution is (6, 2). x=6Simplify. 4x – 3y=18Second equation 4x – 3(2)=18y = 2 4x – 6=18Simplify. 4x – 6 + 6=18 + 6Add 6 to each side. 4x=24Simplify. Divide each side by 4.

26. Example 3 Use elimination to solve the system of equations. 9x – 2y = 30 x – 2y = 14 A.(2, 2) B.(–6, –6) C.(–6, 2) D.(2, –6)

27. Solve the system of equations. 1)8b + 3c = 11 8b + 7c = 7 (1.75, 1) 2) a + 4b = -4 a + 10b = 16 (4, -2) 3) 6c – 9d = 111 5c – 9d = 103 (8, -7)

28. 4) 6x – 2y = 1 10x – 2y = 5 (1, 2.5) 5) -3x - 8y = -24 -3x + 5y = -4.5 (4, 1.5)

29. Example 4 Write and Solve a System of Equations RENTALS A hardware store earned $956.50 from renting ladders and power tools last week. The store charged 36 days for ladders and 85 days for power tools. This week the store charged 36 days for ladders, 70 days for power tools, and earned $829. How much does the store charge per day for ladders and for power tools? UnderstandYou know the number of days the ladders and power tools were rented and the total cost for each.

30. Example 4 Write and Solve a System of Equations PlanLet x = the cost per day for ladders rented and y = the cost per day for power tools rented. LaddersPower ToolsEarnings 36x+85y=956.50 36x+70y= 829 SolveSubtract the equations to eliminate one of the variables. Then solve for the other variable.

31. Example 4 Write and Solve a System of Equations Write the equations vertically. Now substitute 8.5 for y in either equation. 36x+85y=956.50 (–)36x+70y=829 15y=127.5Subtract. y=8.5Divide each side by 15.

32. Example 4 Write and Solve a System of Equations 36x + 85y=956.50First equation 36x + 85(8.5)=956.50Substitute 8.5 for y. 36x + 722.5=956.50Simplify. 36x=234Subtract 722.5 from each side. x=6.5Divide each side by 36. Answer: The store charges $6.50 per day for ladders and $8.50 per day for power tools. Check Substitute both values into the other equation to see if the equation holds true. If x = 6.5 and y = 8.5, then 36(6.5) + 70(8.5) = 829.

33. Cheryl and Jackie work at an ice cream shop. Cheryl earns $8.50 per hour and Jackie earns $7.50 per hour. During a typical wee, Cheryl and Jackie earn $299.50 together. One wee, Jackie doubles her work hours, and the girls earn $412. How many hours does each girl work during a typical week? 8.50c + 7.50j = 299.50 8.50c + (2)(7.50)j = 421 Cheryl works 22 hours, Jackie works 15 hours

34. Use elimination to solve each system of equations 1)4(x + 2y) = 8 4x + 4y = 12 Use the Distributive Property first 4x + 8y = 84x + 8y = 8 (-) 4x + 4y = 12 -4x – 4y = -12 4y = -4 4 4 y = -1

35. Now solve for x, put the value of y in the place of y in the either equation 4(x + 2(-1)) = 8 or 4x + 4(-1) = 12 4(x – 2) = 8 4x – 4 = 12 4x – 8 = 8 +4 +4 +8 +8 4x = 16 4x = 16 4 4 4 4 x = 4 x = 4 (4, -1)

36. 1)3x – 5y = 11 5(x + y) = 5 (2, -1) 2) 1/2x + 2/3y = 2 ¾ 1/4x – 2/3y = 6 ¼ (12, -4 7/8) 3) 3/5x + 1/2y = 8 1/3 -3/5x + 3/4y = 8 1/3 (2 7/9, 13 1/3)

37. Example 4 A.Marcus: $22.00, Anisa: $21.65 B.Marcus: $21.00, Anisa: $22.50 C.Marcus: $24.00, Anisa: $20.00 D.Marcus: $20.75, Anisa: $22.75 FUNDRAISING For a school fundraiser, Marcus and Anisa participated in a walk-a-thon. In the morning, Marcus walked 11 miles and Anisa walked 13. Together they raised $523.50. After lunch, Marcus walked 14 miles and Anisa walked 13. In the afternoon they raised $586.50. How much did each raise per mile of the walk-a-thon?

38. End of the Lesson