1. Solving Systems Using Elimination Section 6-3 Part 2

2. Goals Goal To solve systems by adding or subtracting to eliminate a variable. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary Elimination Method

4. Further Elimination In Part 1 of this lesson, you was that to eliminate a variable, its coefficients must have a sum or difference of zero. In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients, so you can add or subtract to eliminate the variable.

5. Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Multiply the equations and solve. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations. Further Elimination Procedure

6. 2x + 2y = 6 3x – y = 5 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! None of the coefficients are the same! Find the least common multiple of each variable. LCM = 6x, LCM = 2y Which is easier to obtain? 2y (you only have to multiply the bottom equation by 2) Example: Multiplying One Equation

7. Step 4: Plug back in to find the other variable. 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y = 1 2x + 2y = 6 3x – y = 5 Step 3: Multiply the equations and solve. Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 8x = 16 x = 2 2x + 2y = 6 (+) 6x – 2y = 10 Example: Continued

8. Step 5: Check your solution. (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 2x + 2y = 6 3x – y = 5 Solving with multiplication adds one more step to the elimination process. Example: Continued

9. x + 4y = 7 4x – 3y = 9 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. Find the least common multiple of each variable. LCM = 4x, LCM = 12y Which is easier to obtain? 4x (you only have to multiply the top equation by -4 to make them inverses) Example: Multiplying One Equation

10. x + 4y = 7 4x – 3y = 9 Step 4: Plug back in to find the other variable. x + 4(1) = 7 x + 4 = 7 x = 3 Step 3: Multiply the equations and solve. Multiply the top equation by -4 (-4)(x + 4y = 7) 4x – 3y = 9) y = 1 -4x – 16y = -28 (+) 4x – 3y = 9 -19y = -19 Example: Continued

11. Step 5: Check your solution. (3, 1) (3) + 4(1) = 7 4(3) - 3(1) = 9 x + 4y = 7 4x – 3y = 9 Example: Continued

12. x + 2y = 11 –3x + y = –5 Solve the system by elimination. Multiply each term in the second equation by –2 to get opposite y-coefficients. x + 2y = 11 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) Add the new equation to the first equation. 7x + 0 = 21 7x = 21 x = 3 Simplify and solve for x. Your Turn:

13. Write one of the original equations. x + 2y = 11 Substitute 3 for x. 3 + 2y = 11 Subtract 3 from each side. –3 2y = 8 y = 4 Simplify and solve for y. Write the solution as an ordered pair. (3, 4) Continued

14. Solve the system by elimination. 3x + 2y = 6 –x + y = –2 3x + 2y = 6 3(–x + y = –2) 3x + 2y = 6 +(–3x + 3y = –6) 0 + 5y = 0 Multiply each term in the second equation by 3 to get opposite x-coefficients. Add the new equation to the first equation. Simplify and solve for y. 5y = 0 y = 0 Your Turn:

15. Write one of the original equations. –x + y = –2 Substitute 0 for y. –x + 3(0) = –2 –x + 0 = –2 –x = –2 Simplify and solve for x. Write the solution as an ordered pair. (2, 0) x = 2 Continued

16. 3x + 4y = -1 4x – 3y = 7 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. Find the least common multiple of each variable. LCM = 12x, LCM = 12y Which is easier to obtain? Either! I’ll pick y because the signs are already opposite. Example: Multiplying Both Equations

17. 3x + 4y = -1 4x – 3y = 7 Step 4: Plug back in to find the other variable. 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 Step 3: Multiply the equations and solve. Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) x = 1 9x + 12y = -3 (+) 16x – 12y = 28 25x = 25 Example: Continued

18. Step 5: Check your solution. (1, -1) 3(1) + 4(-1) = -1 4(1) - 3(-1) = 7 3x + 4y = -1 4x – 3y = 7 Example: Continued

19. –5x + 2y = 32 2x + 3y = 10 Solve the system by elimination. 2(–5x + 2y = 32) 5(2x + 3y = 10) Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients –10x + 4y = 64 +(10x + 15y = 50) Add the new equations. Simplify and solve for y. 19y = 114 y = 6 Your Turn:

20. Write one of the original equations. 2x + 3y = 10 Substitute 6 for y. 2x + 3(6) = 10 Subtract 18 from both sides. –18 2x = –8 2x + 18 = 10 x = –4 Simplify and solve for x. Write the solution as an ordered pair. (–4, 6) Continued

21. Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25 3(2x + 5y = 26) +(2)(–3x – 4y = –25) Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients 6x + 15y = 78 +(–6x – 8y = –50) Add the new equations. Simplify and solve for y. y = 4 0 + 7y = 28 Your Turn:

22. Write one of the original equations. 2x + 5y = 26 Substitute 4 for y. 2x + 5(4) = 26 Simplify and solve for x. Write the solution as an ordered pair. (3, 4) x = 3 2x + 20 = 26 –20 2x = 6 Subtract 20 from both sides. Your Turn:

23. N UMBER OF S OLUTIONS OF A L INEAR S YSTEM IDENTIFYING THE NUMBER OF SOLUTIONS C ONCEPT S UMMARY y x y x Lines intersect one solution Lines are parallel no solution y x Lines coincide infinitely many solutions

24. Identifying The Number of Solutions If both variable terms are eliminated as you solve a system of equations, the answer is either no solution or infinite solutions. –No solution: get a false statement when solving the system. –Infinite solutions: get a true statement when solving the system.

25. Show that this linear system has infinitely many solutions. – 2 x  y  3 Equation 1 – 4 x  2y  6 Equation 2 M ETHOD : Elimination You can multiply Equation 1 by –2. 4x – 2y  – 64x – 2y  – 6 Multiply Equation 1 by –2. – 4 x  2y  6 Write Equation 2. 0  0 Add Equations. True statement! The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This result tells you that the linear system has infinitely many solutions. A Linear System with Infinite Solutions

26. Show that this linear system has no solution. 2 x  y  5 Equation 1 2 x  y  1 Equation 2 You can multiply Equation 1 by –1. -2 x - y  -5 Multiply Equation 1 by –1. 2 x  y  1 Write Equation 2. 0  - 4 Add Equations. False statement! The variables are eliminated and you are left with a statement that is false regardless of the values of x and y. This result tells you that the linear system has no solution. A Linear System with No Solution M ETHOD : Elimination

27. Your Turn: Solve the systems using elimination. 1) 2) False Statement No Solution True Statement Infinite Solutions

28. Summary

29. Summary of Methods for Solving Systems 7.3 The Elimination Method Substitution The value of one variable is known and can easily be substituted into the other equation. 6x + y = 10 y = 5 Example Suggested Method Why

30. Summary of Methods for Solving Systems 7.3 The Elimination Method Elimination eliminate ‘y’  5 Add the two equations 2x – 5y = –20 4x + 5y = 14 Example Suggested Method Why

31. Summary of Methods for Solving Systems 7.3 The Elimination Method Elimination 9a – 2b = –11 8a + 4b = 25 Example Suggested Method Why eliminate ‘b’  4 Multiply first equation by 2 Add the equations

32. Joke Time How does an octopus go to war? Well-Armed! Why does a Moo-rock taste better than an Earth-rock? Because it’s a little meteor! What did the elder chimney say to the younger chimney? You’re too young to smoke!

33. Assignment 6-3 Part 2 Exercises Pg. 402 - 404: #6 – 34 even