1. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §3.3a 3-Var Linear Systems

2. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §3.2b → More System Applications  Any QUESTIONS About HomeWork §3.2 → HW-09 3.2 MTH 55

3. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 3 Bruce Mayer, PE Chabot College Mathematics Systems of 3-Variables  A linear equation in three variables is an equation equivalent to form Ax + By + Cz = D Where A, B, C, and D are real numbers and A, B, and C are not all 0  A solution of a system of three equations in three variables is an ORDERED TRIPLE (x 1, y 1, z 1 ) that makes all three equations true.

4. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 4 Bruce Mayer, PE Chabot College Mathematics Example  Ordered Triple Soln  Determine whether (2, −1, 3) is a solution of the system  SOLUTION: In all three equations, replace x with 2, y with −1, and z with 3. x + y + z = 4 2 + (–1) + 3 = 4 4 = 43 = 3 2x – 2y – z = 3 2(2) – 2(–1) – 3 = 3 – 4x + y + 2z = –3 – 4(2) + (–1) + 2(3) = –3 –3 = –3 

5. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Ordered Triple Soln  Determine whether (2, −1, 3) is a solution of the system  SOLUTION: Because (2, −1, 3) satisfies all three equations in the system, this triple IS a solution for the system.

6. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 6 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 1.Write each equation in the standard form of Ax + By+ Cz = D. 2.Eliminate one variable from one PAIR of equations using the elimination method. 3.If necessary, eliminate the same variable from another PAIR of equations to produce a system of TWO varialbles in TWO Equations

7. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 7 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 4.Steps 2 and 3 result in two equations with the same two variables. Solve these equations using the elimination method. 5.To find the third variable, substitute the values of the variables found in step 4 into any of the three original equations that contain the 3 rd variable 6.Check the ordered triple in all three of the original equations

8. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Solve System of Equations (1) (2) (3)  SOLUTION: select any two of the three equations and work to get one equation in two variables. Let’s add eqs (1) & (2) (1) (2) (4) 2x + 3y = 8 Adding to eliminate z

9. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 9 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Next, select a different pair of equations and eliminate the same variable. Use (2) & (3) to again eliminate z. (5) x – y + 3z = 8 Multiplying equation (2) by 3 4x + 5y = 14. 3x + 6y – 3z = 6

10. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 10 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Now solve the resulting system of eqns (4) & (5). That will give us two of the numbers in the solution of the original system  Multiply both sides of eqn (4) by −2 and then add to eqn (5): 4x + 5y = 14 2x + 3y = 8 (5) (4) 4x + 5y = 14 –4x – 6y = –16, –y = –2 y = 2

11. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 11 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Substituting into either equation (4) or (5) we find that x = 1  Now we have x = 1 and y = 2. To find the value for z, we use any of the three original equations and substitute to find the third number z.  Let’s use eqn (1) and substitute our two numbers in it: x + y + z = 6 1 + 2 + z = 6 z = 3.

12. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  At this point we have solved for All Three variables in the System: z = 3y = 2x = 1  Thus We have obtained the ordered triple (1, 2, 3). It should be checked in all three equations  Finally the Solution Set: (1, 2, 3)

13. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Solve System of Equations  SOLUTION: The equations are in standard form and do not contain decimals or fractions.  Eliminate z from eqns (2) & (3). (1) (2) (3) (2) (3) (4) 3x + 2y = 4

14. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Eliminate z from equations (1) and (2).  Eliminate x using above and eqn (4) Multiplying equation (2) by 6 Adding Eqns 15x + 15y = 15 3x + 2y = 4 15x + 15y = 15 Multiplying top by  5  15x – 10y =  20 15x + 15y = 15 Adding Eqns 5y =  5 y =  1

15. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 15 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination  Using y = −1, find x from equation 4 by substituting  Substitute x = 2 and y = −1 to find z 3x + 2y = 4 3x + 2(  1) = 4 x = 2 x + y + z = 2 2 – 1 + z = 2 1 + z = 2 z = 1  Thus The solution is the ordered triple (2, −1, 1)

16. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination x – 4 y + 3 z = –9(3) Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 1 Eliminate a variable from the sum of two equations. The choice of the variable to eliminate is arbitrary. We will eliminate z. 6 x + 9 y – 3 z = 15 7 x + 5 y = 6Add. (4) Multiply each side of (1) by 3. 7 x + 5 y = 6 (4)

17. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination 4 x – 16 y + 12 z = –36 Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 2 Eliminate the same variable, z, from any other equations. –9 x + 6 y – 12 z = 6 –5 x – 10 y = –30 Add. (5) Multiply each side of (2) by 3. 7 x + 5 y = 6 (4) Multiply each side of (3) by 4. –5 x – 10 y = –30 (5)

18. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination –5 x – 10 y = –30 Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 3 Eliminate a different variable and solve. 14 x + 10 y = 12 9 x = –18 Add. (6) Multiply each side of (4) by 2. 7 x + 5 y = 6 (4) (5) –5 x – 10 y = –30 (5) x = –2

19. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 4 Find a second value by substituting –2 for x in (4) or (5). 7 x + 5 y = 6(4) 7 x + 5 y = 6 (4) Recall x = –2. –5 x – 10 y = –30 (5) y = 4 7(–2) + 5 y = 6 –14 + 5 y = 6 5 y = 20

20. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 5 Find a third value by substituting –2 for x and 4 for y into any of the three original equations. 2 x + 3 y – z = 5(1) 7 x + 5 y = 6 (4) From Before x = –2 and y = 4 –5 x – 10 y = –30 (5) 2(–2) + 3(4) – z = 5 8 – z = 5 z = 3

21. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Elimination Solve the system. 2 x + 3 y – z = 5(1) –3 x + 2 y – 4 z = 2(2) x – 4 y + 3 z = –9(3) Step 6 Check that the ordered triple (–2, 4, 3) is the solution of the system. The ordered triple must satisfy all three original equations. 2 x + 3 y – z = 5 2(–2) + 3(4) – 3 = 5 –4 + 12 – 3 = 5 5 = 5  (1) –3 x + 2 y – 4 z = 2 –3(–2) + 2(4) – 4(3) = 2 6 + 8 – 12 = 2 2 = 2  (2) x – 4 y + 3 z = –9 (–2) – 4(4) + 3(3) = –9 –2 – 16 + 9 = –9 –9 = –9  (3)

22. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Missing Term(s) Multiply each side of (2) by 4. –20 y – 8 z = 12 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Since equation (3) is missing the variable y, a good way to begin the solution is to eliminate y again by using equations (1) and (2). 25 x + 20 y = 115 25 x – 8 z = 127 Add. (4) 25 x – 8 z = 127 (4) Multiply each side of (1) by 5.

23. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 23 Bruce Mayer, PE Chabot College Mathematics Example  Missing Term(s) Multiply each side of (4) by 9. 225 x – 72 z = 1143 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Now use equations (3) and (4) to eliminate z and solve. –64 x + 72 z = –16 161 x = 1127 Add. (5) 25 x – 8 z = 127 (4) Multiply each side of (3) by 8. x = 7

24. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Missing Term(s) 5(7) + 4 y = 23 Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3) Substituting into equation (1) gives 5 x + 4 y = 23 35 + 4 y = 23 25 x – 8 z = 127 (4) 4 y = –12 y = –3. –5(–3) – 2 z = 3 Substituting into equation (2) gives –5 y – 2 z = 3 15 – 2 z = 3 –2 z = –12 z = 6.

25. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Missing Term(s) Solve the system. 5 x + 4 y = 23 (1) –5 y – 2 z = 3 (2) –8 x + 9 z = –2 (3)  Thus, x = 7, y = −3, and z = 6. Check these values in each of the original equations of the system to verify that the solution set of the system is the triple (7, −3, 6)

26. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 26 Bruce Mayer, PE Chabot College Mathematics Inconsistent System  If, in the process of performing elimination on a linear system, an equation of the form 0 = a occurs, where a ≠ 0, then the system has no solution and is inconsistent.

27. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  Inconsistent Sys  Solve the System of Equations  SOLUTION: To eliminate x from Eqn (2), add −2 times Eqn (1) to Eqn (2)

28. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example  Inconsistent Sys  Next add −3 times Eqn (1) to Eqn (3) to eliminate x from Eqn (3)  We now have the following system

29. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example  Inconsistent Sys  Now Multiply Eqn (4) by 1/3 to obtain  To eliminate y from Eqn (5), add −1 times Eqn (6) to Eqn (5)

30. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 30 Bruce Mayer, PE Chabot College Mathematics Example  Inconsistent Sys  We now have the system in simplified elimination form:  This system is equivalent to the original system. Since equation (7) is false (a contradiction), we conclude that the solution set of the system is Ø (the NULL Set), and the system is INconsistent.

31. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 31 Bruce Mayer, PE Chabot College Mathematics Dependent Systems  If, in the process of performing an elimination solution for a linear system i.an equation of the form 0 = a (a ≠ 0) does not occur, but ii.an equation of the form 0 = 0 does occur, then the system of equations has infinitely many solutions and the equations are dependent.

32. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 32 Bruce Mayer, PE Chabot College Mathematics Example  Dependent System  Solve the System of Equations  SOLUTION: Eliminate x from Eqn (2) by adding −3 times Eqn (1) to Eqn (2

33. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example  Dependent System  Eliminate x from Eqn (3) by adding −4 times Eqn (1) to Eqn (3)  We now have the equivalent system

34. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  Dependent System  To eliminate y from Eq (5), add −1 times Eqn (4) to Eqn (5)  We now have the equivalent system in Final Form

35. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 35 Bruce Mayer, PE Chabot College Mathematics Example  Dependent System  The equation 0 = 0 may be interpreted as 0·z = 0, which is true for every value of z. Solving eqn (4) for y, we have y = 3z – 2. Substituting into eqn (1) and solving for x.  Thus every triple (x, y, z) = (2z + 5, 3z – 2, z) is a soln of the system for each value of z. E.g, for z = 1, the triple is (7, 1, 1).

36. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 36 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems  The graph of a linear equation in three variables, such as Ax + By + Cz = C (where A, B, and C are not all zero), is a plane in three-Dimensional (3D) space.  Following are the possible situations for a system of three linear equations in three variables.

37. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 37 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems a)Three planes intersect in a single point. The system has only one solution b)Three planes intersect in one line. The system has infinitely many solutions

38. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 38 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems c)Three planes coincide with each other. The system has only one solution. d)There are three parallel planes. The system has no solution.

39. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 39 Bruce Mayer, PE Chabot College Mathematics Geometry of 3Var Systems e)Two parallel planes are intersected by a third plane. The system has no solution. f)Three planes have no point in common. The system has no solution.

40. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 40 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Modeling  The following table shows the higher-order multiple birth rates in the USA since 1971. At the right is a scatter plot of this data.

41. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 41 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Modeling  We will SELECT three arbitrary ordered pairs to construct the model  We Pick the Points (1,29), (11,40) and (21,100) and substitute the x & y values from these ordered pairs into the Quadratic Function

42. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 42 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Modeling  Selecting three representative ordered pairs, we can write a system of three equations.  Solving this 3-Variable System by Elimination:  This produces our Model

43. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 43 Bruce Mayer, PE Chabot College Mathematics Example  Quadratic Modeling  Use to the Model to Estimate the Hi- Order Births in 1993  In 1993 x = 23  Sub 23 into Model  Thus we estimate that in 1993 the Hi-Order BirthRate was about 118

44. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 44 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §3.3 Exercise Set 28  Quadratic Function

45. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 45 Bruce Mayer, PE Chabot College Mathematics All Done for Today A CONSISTENT System z = -1.553x - 2.642y - 10.272 (darker green) z = 1.416x - 1.92y - 10.979 (medium green) z = -.761x -.236y - 7.184 (lighter green) THE THREE PLANES SHARE ONE POINT

46. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 46 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –

47. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 47 Bruce Mayer, PE Chabot College Mathematics Equivalent Systems  Operations That Produce Equivalent Systems 1.Interchange the position of any two eqns 2.Multiply any eqn by a nonzero constant 3.Add a nonzero multiple of one eqn to another  A special type of Elimination called Gaussian Elimination uses these steps to solve multivariable systems

48. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 48 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 1.Rearrange the equations, if necessary, to obtain an x-term with a nonzero coefficient in the first equation. Then multiply the first equation by the reciprocal of the coefficient of the x-term to get 1 as a leading coefficient. 2.By adding appropriate multiples of the first equation, eliminate any x-terms from the second and third equations

49. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 49 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 2.(cont.) Multiply the resulting second equation by the reciprocal of the coefficient of the y-term to get 1 as the leading coefficient. 3.If necessary by adding appropriate multiple of the second equation from Step 2, eliminate any y-term from the third equation. Solve the resulting equation for z.

50. BMayer@ChabotCollege.edu MTH55_Lec-13_sec_3-3a_3Var_Lin_Sys.ppt 50 Bruce Mayer, PE Chabot College Mathematics Solve 3Var Sys by Elimination 4.Back-substitute the values of z from Steps 3 into one of the equations in Step 3 that contain only y and z, and solve for y. 5.Back-substitute the values of y and z from Steps 3 and 4 in any equation containing x, y, and z, and solve for x 6.Write the solution set (Soln Triple) 7.Check soln in the original equations