1. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics §3.1 2-Var Linear Systems

2. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §’s2.4 → Point-Slope Eqn, Modeling  Any QUESTIONS About HomeWork §’s2.4 → HW-07 2.4 MTH 55

3. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 3 Bruce Mayer, PE Chabot College Mathematics Systems of Equations  System of Equations ≡ A group of two or more equations; e.g.,  Solution For A System Of Equations ≡ An ordered set of numbers that makes ALL equations in the system TRUE at the same time

4. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 4 Bruce Mayer, PE Chabot College Mathematics Checking System Solution  To verify or check a solution to a system of equations: 1.Replace each variable in each equation with its corresponding value. 2.Verify that each equation is true.

5. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Chk System Soln  Consider The Equation System  Determine whether each ordered pair is a solution to the system of equations. a. (−3, 2)b. (3, 4)

6. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Chk System Soln  SOLUTION → Chk True/False  a. (−3, 2) → Sub: −3 for x, & 2 for y x + y = 7y = 3x − 2 −3 + 2 = 7 2 = 3(−3) − 2 −1 = 72 = −11 FalseFalse

7. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 7 Bruce Mayer, PE Chabot College Mathematics Example  Chk System Soln  SOLUTION → Chk True/False  b. (3, 4) → Sub: 3 for x, & 4 for y x + y = 7y = 3x − 2 3 + 4 = 7 4 = 3(3) − 2 7 = 74 = 7 TrueFalse

8. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 8 Bruce Mayer, PE Chabot College Mathematics Example  Chk System Soln  SOLUTION → Chk True/False  Because (−3, 2) does NOT satisfy EITHER equation, it is NOT a solution for the system.  Because (3, 4) satisfies ONLY ONE equation, it is NOT a solution to the system of equations

9. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 9 Bruce Mayer, PE Chabot College Mathematics Systems of Equations Soln  A system-of- equations problem involves finding the solutions that satisfy the conditions set forth in two or more Equations  For Equations of Lines, The System Solution is the CROSSING Point  This Graph shows two lines which have one point in common  The common point is (–3,2) Satisfies BOTH Eqns

10. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 10 Bruce Mayer, PE Chabot College Mathematics Solve Systems of Eqns by Graphing  Recall that a graph of an equation is a set of points representing its solution set  Each point on the graph corresponds to an ordered pair that is a solution of the equation  By graphing two equations using one set of axes, we can identify a solution of both equations by looking for a point of intersection

11. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 11 Bruce Mayer, PE Chabot College Mathematics Solving by Graphing Procedure 1.Write the equations of the lines in slope-intercept form. 2.Use the slope and y-intercept of each line to plot two points for each line on the same graph. 3.Draw in each line on the graph. 4.Determine the point of intersection (the Common Pt) and write this point as an ordered pair for the Solution

12. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 12 Bruce Mayer, PE Chabot College Mathematics Example  Solve w/ Graphing  Solve this system y = 3x + 1 x – 2y = 3  SOLUTION: Graph Each Eqn y = 3x + 1 –Graph (0, 1) and “count off” a slope of 3 x – 2y = 3 –Graph using the intercepts: (0,–3/2) & (3, 0) (−1, −2) The crossing point provides the common solution

13. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Solve w/ Graphing  Chk (−1, −2) Soln: y = 3x + 1 x − 2y = 3  y = 3x + 1→ −2 = 3(−1) + 1 −2 = −3 + 1 −2 = −2   x − 2y = 3 → (−1) − 2(−2) = 3 −1+4 = 3 3 = 3   Thus (−1, −2) Chks as a Soln

14. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Solve By Graphing  Solve System:  SOLUTION: graph Both Equations As a check note that [4−2] = [6−4] is true. –The solution is (4, 2) The graphs intersect at (4, 2), indicating that for the x-value 4 both x−2 and 6−x share the same value (in this case 2). (4, 2) y = x  2 y = 6  x

15. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 15 Bruce Mayer, PE Chabot College Mathematics The Substitution Soln Method  Graphing can be an imprecise method for solving systems of equations.  We are now going to look at ways of finding exact solutions using algebra  One method for solving systems is known as the substitution method. It uses algebra instead of graphing and is thus considered an algebraic method.

16. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 16 Bruce Mayer, PE Chabot College Mathematics Substitution Summarized  The substitution method involves isolating either variable in one equation and substituting the result for the same variable in the second equation. The numerical result is then back-substituted into the first equation to find the numerical result for the second variable

17. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Subbing  Solve the System  SOLUTION: The second equation says that y and −2x + 5 represent the same value.  Thus, in the first equation we can substitute −2x + 5 for y

18. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Subbing  The Algebra to Solve Equation (1) Substitute: y = −2x + 5 Distributive Property Combine Like Terms Add 10 to Both Sides Divide Both Sides by 7 to Find x

19. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Subbing  We have found the x-value of the solution. To find the y-value, we return to the original pair of equations. Substituting x=2 into either equation will give us the y-value. Choose eqn (2): Equation (2) Substitute: x = 2 Simplifying When x = 2  The ordered pair (2, 1) appears to be the solution

20. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Subbing  Check Tentative Solution (2,1) 3x − 2y = 4 y = −2x + 5 3(2) − 2(1) 4 1 −2(2) + 5 6 − 2 4 1 −4 + 5 4 = 4 True 1 = 1 True  Since (2, 1) checks in BOTH equations, it IS a solution.

21. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 21 Bruce Mayer, PE Chabot College Mathematics Substitution Solution CAUTION  Caution  Caution! A solution of a system of equations in two variables is an ordered pair of numbers. Once you have solved for one variable, do not forget the other. A common mistake is to solve for only one variable.

22. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Subbing  Solve the System  SOLUTION: Sub 3 − y for x in Eqn (2) Equation (2) Substitute: x = 3−y Distributive Property Combine Terms, Subtract 5 from Both sides Solve for y = 5

23. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 23 Bruce Mayer, PE Chabot College Mathematics Solve by Subbing  Find x for y = 5 Use Eqn (1) Eqn (1) Sub y = 5 Solve for x  Chk Soln pair (−2,5) x = 3 − y 5x + 3y = 5 −2 = 3 − 5 5(−2)+3(5) = 5 −2 = −2 −10 + 15 = 5 5 = 5  Thus (−2,5) is the Soln The graph below is another check. ( − 2, 5) x = 3  y 5x + 3y = 5

24. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 24 Bruce Mayer, PE Chabot College Mathematics Solving for the Variable First  Sometimes neither equation has a variable alone on one side. In that case, we solve one equation for one of the variables and then proceed as before  For Example; Solve:  We can solve either equation for either variable. Since the coefficient of x is one in equation (1), it is easier to solve that equation for x.

25. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Solve  substitute x = 6−y for x in equation (2) of the original pair and solve for y Equation (2) Substitute: x = 6−y Distributive Property Combine Like Terms Add −8, Add 3y to Both Sides Divide Both Sides by 3 to Find y Use Parens or Brackets when Subbing

26. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Solve  To find x we substitute 22/3 for y in equation (1), (2), or (3). Because it is generally easier to use an equation that has already been solved for a specific variable, we decide to use equation (3):  A Check of this Ordered Pair Shows that it is a Solution:

27. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 27 Bruce Mayer, PE Chabot College Mathematics Substitution Solution Procedure 1.Solve for a variable in either one of the equations if neither equation already has a variable isolated. 2.Using the result of step (1), substitute in the other equation for the variable isolated in step (1). 3.Solve the equation from step (2). 4.Substitute the ½-solution from step (3) into one of the other equations to solve for the other variable. 5.Check that the ordered pair resulting from steps (3) and (4) checks in both of the original equations.

28. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 28 Bruce Mayer, PE Chabot College Mathematics Solving by Addition/Elimination  The Addition/Elimination method for solving systems of equations makes use of the addition principle  For Example; Solve System same thing  According to equation (2), x−3y and 7 are the same thing. Thus we can add 4x + 3y to the left side of the equation(1) and 7 to the right side

29. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 29 Bruce Mayer, PE Chabot College Mathematics Elimination Example  Add Equations  The resulting equation has just one variable: 5x = 15 Dividing both sides by 5, find that x = 3  Next Sub 3 for x in Either Eqn to Find the y-value Using Eqn-1

30. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 30 Bruce Mayer, PE Chabot College Mathematics Elimination Example  Check Tentative Solution (3, −4/3) Since (3, −4/3) checks in both equations, it is the solution –Graph confirms 4x + 3y = 8 x − 3y = 7 4(3) + 3(−4/3) 8 3 − 3(−4/3) 7 12 − 4 3 + 4 8 = 8 7 = 7True

31. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 31 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: Adding the two equations as they appear will not eliminate a variable.  However, if the 3y were −3y in one equation, we could eliminate y. We multiply both sides of equation (2) by −1 to find an equivalent eqn and then add:

32. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 32 Bruce Mayer, PE Chabot College Mathematics Example  Solve  To Find y substitute 2 for x in either of the original eqns:  The graph shown below also checks.  We can check the ordered pair (2, 3) in Both Eqns 

33. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 33 Bruce Mayer, PE Chabot College Mathematics Which Variable to Eliminate  When deciding which variable to eliminate, we inspect the coefficients in both equations. If one coefficient is a simple multiple of the coefficient of the same variable in the other equation, that one is the easiest variable to eliminate.  For Example; Solve System

34. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: No terms are opposites, but if both sides of equation (1) are multiplied by 2, the coefficients of y will be opposites. Mult Both Sides of Eqn-1 by 2 Add Eqns Solve for x

35. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 35 Bruce Mayer, PE Chabot College Mathematics Example  Solve  SOLUTION: Find y by Subbing x=2 into Either Eqn; Choosing Eqn-1 Sub 2 for x Simplify Solve for y  Student Exercise: confirm that (2, −1) checks and is the solution.

36. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 36 Bruce Mayer, PE Chabot College Mathematics Multiple Multiplication  Sometimes BOTH equations must be multiplied to find the Least Common Multiple (LCM) of two coefficients  For Example; Solve System  SOLUTION: It is often helpful to write both equations in Standard form before attempting to eliminate a variable:

37. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 37 Bruce Mayer, PE Chabot College Mathematics Example  Solve  Since neither coefficient of x is a multiple of the other and neither coefficient of y is a multiple of the other, we use the multiplication principle with both equations.  We can eliminate the x term by multiplying both sides of equation (3) by 3 and both sides of equation (4) by −2

38. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 38 Bruce Mayer, PE Chabot College Mathematics Example  Solve  The Algebra  Solve for x using y = −2 in Eqn-3 Mult Both Sides of Eqn-3 by 3 Add Eqns Mult Both Sides of Eqn-4 by −2  Students to Verify that solution (−2, −2) checks

39. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 39 Bruce Mayer, PE Chabot College Mathematics Solving Systems by Elimination 1.Write the equations in standard form (Ax + By = C). 2.Use the multiplication principle to clear fractions or decimals. 3.Multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. 4.Add the equations. The result should be an equation in terms of one variable. 5.Solve the equation from step 4 for the value of that variable. 6.Using an equation containing both variables, substitute the value you found in step 5 for the corresponding variable and solve for the value of the other variable. 7.Check your solution in the original equations

40. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 40 Bruce Mayer, PE Chabot College Mathematics Types of Systems of Equations  When solving problems concerning systems of two linear equations and two variables there are three possible outcomes. 1.Consistent Systems 2.INConsistent Systems 3.Dependent Systems

41. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 41 Bruce Mayer, PE Chabot College Mathematics Case-1 Consistent Systems  In this case, the graphs of the two lines intersect at exactly one point.

42. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 42 Bruce Mayer, PE Chabot College Mathematics Case-2 Inconsistent Systems  In this case the graphs of the two lines show that they are parallel.  Since there is NO Intersection, there is NO Solution to this system

43. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 43 Bruce Mayer, PE Chabot College Mathematics Case-3 Dependent Systems  In this case the graphs of the two lines indicate that there are infinite solutions because they are, in reality, the same line.

44. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 44 Bruce Mayer, PE Chabot College Mathematics Testing for System Case  Given System of Two Eqns  Case-1 Consistent → m 1 ≠ m 2  Case-2 Inconsistent → m 1 = m 2 b 1 ≠ b 2  Case-3 Dependent → m 1 = m 2 AND b 1 = b 2; –Or for a constant K: m 1 = Km 2 AND b 1 = Kb 2

45. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 45 Bruce Mayer, PE Chabot College Mathematics Example  Solve by Graphing  Solve System: Thus the system is INCONSISTENT and has NO solution.  SOLUTION: Graph Eqns equations are in slope- intercept form so it is easy to see that both lines have the same slope. The y-intercepts differ so the lines are parallel. Because the lines are parallel, there is NO point of intersection. y = 3x/4 + 2 y = 3x/4  3

46. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 46 Bruce Mayer, PE Chabot College Mathematics Example  Solve System by Sub  Solve System  SOLUTION: This system is from The previous Graphing example. The lines are parallel and the system has no solution. Let’s see what happens if we try to solve this system by substitution

47. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 47 Bruce Mayer, PE Chabot College Mathematics Example  Solve System by Sub  Solve by Algebra Equation (1) Substitute: y = (3/4)x–4 Subtract (3/4)x from Both Sides  We arrive at contradiction; the system is inconsistent and thus has NO solution

48. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 48 Bruce Mayer, PE Chabot College Mathematics Example  Solve By Graphing  Solve System:  SOLUTION: graph Both Equations equation is a solution of the OTHER equation as well. Both equations represent the SAME line. Because the equations are EQUIVALENT, any solution of ONE

49. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 49 Bruce Mayer, PE Chabot College Mathematics Example  Solve By Subbing  Solve System  SOLUTION: Notice that Eqn-1 is simply TWICE Eqn-2. Thus these Eqns are DEPENDENT, and will Graph as COINCIDENT Lines Recall that Dependent Equations have an INFINITE number of Solutions  Checking by Algebra

50. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 50 Bruce Mayer, PE Chabot College Mathematics Example  Solve By Subbing  Solve by Algebra Equation (1) Use Eqn-2 to Substitute for y Simplifying ReArranging Final ReArrangement

51. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 51 Bruce Mayer, PE Chabot College Mathematics Example  Solve By Subbing  Examine Soln to  This Solution equation is true for any choice of x. When the solution leads to an equation that is true for all real numbers, we state that the system has an infinite number of solutions. Simplification to 0=0 → infinite solns

52. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 52 Bruce Mayer, PE Chabot College Mathematics Example  Elim/Addition Soln  Example – Solve:  SOLUTION: To eliminate y multiply equation (2) by −1. Then add ?¿?¿  Note that in eliminating y, we eliminated x as well. The resulting equation 0 = 4, is false (a contradiction) for any pair (x, y), so there is no solution The Lines are thus PARALLEL

53. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 53 Bruce Mayer, PE Chabot College Mathematics Example  Elim/Addition Soln  Example – Solve:  SOLUTION: To eliminate x, we multiply both sides of eqn (1) by −3, and then add the two eqns ?¿?¿ dependent  Again, we have eliminated both variables. The resulting equation, 0 = 0, is always true, indicating that the equations are dependent

54. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 54 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §3.1 Exercise Set 92, 94, 96  InConsistent Equations x y

55. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 55 Bruce Mayer, PE Chabot College Mathematics All Done for Today Women In The Professions

56. BMayer@ChabotCollege.edu MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt 56 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Chabot Mathematics Appendix –