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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Chabot Mathematics. §3.1 2-Var Linear Systems. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. MTH 55. 2.4. Review §. Any QUESTIONS About §’s2.4 → Point-Slope Eqn, Modeling Any QUESTIONS About HomeWork §’s2.4 → HW-07. Systems of Equations.

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Chabot Mathematics §3.1 2-VarLinear Systems Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. MTH 55 2.4 Review § • Any QUESTIONS About • §’s2.4 → Point-Slope Eqn, Modeling • Any QUESTIONS About HomeWork • §’s2.4 → HW-07

  3. 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. Checking System Solution • To verify or check a solution to a system of equations: • Replace each variable in each equation with its corresponding value. • Verify that each equation is true.

  5. Example  Chk System Soln • Consider TheEquation System • Determine whether each ordered pair is a solution to the system of equations. a. (−3, 2) b. (3, 4)

  6. Example  Chk System Soln • SOLUTION →Chk True/False • a. (−3, 2) → Sub: −3 for x, & 2 for y x + y = 7 y = 3x − 2 −3 + 2 = 7 2 = 3(−3) − 2 −1 = 7 2 = −11 FalseFalse

  7. Example  Chk System Soln • SOLUTION →Chk True/False • b. (3, 4) → Sub: 3 for x, & 4 for y x + y = 7 y = 3x − 2 3 + 4 = 7 4 = 3(3) − 2 7 = 7 4 = 7 TrueFalse

  8. 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. 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 Systems of Equations Soln • This Graph shows two lines which have one point in common • The common point is (–3,2) Satisfies BOTH Eqns

  10. 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. Solving by Graphing Procedure • Write the equations of the lines in slope-intercept form. • Use the slope and y-intercept of each line to plot two points for each line on the same graph. • Draw in each line on the graph. • Determine the point of intersection (the Common Pt) and write this point as an ordered pair for the Solution

  12. 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) Example  Solve w/ Graphing (−1, −2) The crossing point provides the common solution

  13. Chk (−1, −2) Soln: y = 3x + 1 x −2y = 3 Example  Solve w/ Graphing • 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) Chksas a Soln

  14. Solve System: Example  Solve By Graphing y = 6  x (4, 2) • SOLUTION: graph Both Equations • 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). y = x  2 • As a check note that [4−2] = [6−4] is true. • The solution is (4, 2)

  15. 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. 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. Example  Solve by Subbing • Solve theSystem • 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. Example  Solve by Subbing • The Algebrato 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. 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) • The ordered pair (2, 1) appears to be the solution Substitute: x = 2 Simplifying When x = 2

  20. 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. Substitution Solution 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. Example  Solve by Subbing • Solve theSystem • 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. Find x for y = 5 Use Eqn (1) Solve by Subbing • Thus (−2,5) is the Soln • The graph below is another check. Eqn (1) 5x + 3y = 5 Sub y = 5 Solve for x (−2, 5) • Chk Soln pair (−2,5) x = 3  y x = 3 −y5x + 3y = 5 −2 = 3 −5 5(−2)+3(5) = 5 −2 = −2 −10 + 15 = 5 5 = 5

  24. 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: Solving for the Variable First • 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. 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 Use Parens or Brackets when Subbing Distributive Property Combine Like Terms Add −8, Add 3y to Both Sides Divide Both Sides by 3 to Find y

  26. 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. Substitution Solution Procedure • Solve for a variable in either one of the equations if neither equation already has a variable isolated. • Using the result of step (1), substitute in the other equation for the variable isolated in step (1). • Solve the equation from step (2). • Substitute the ½-solution from step (3) into one of the other equations to solve for the other variable. • Check that the ordered pair resulting from steps (3) and (4) checks in both of the original equations.

  28. Solving by Addition/Elimination • The Addition/Elimination method for solving systems of equations makes use of the addition principle • For Example;Solve System • According to equation (2), x−3y and 7 are the samething. Thus we can add 4x + 3y to the left side of the equation(1) and 7 to the right side

  29. Elimination Example • AddEquations • The resulting equation has just one variable: 5x = 15 • Dividing both sides by 5, find that x = 3 • Next Sub 3 for x inEither Eqn to Findthe y-value • Using Eqn-1

  30. Elimination Example • Check Tentative Solution (3, −4/3) 4x + 3y = 8x− 3y = 7 4(3) + 3(−4/3) 8 3− 3(−4/3) 7 12 − 4 3 + 4 8 = 8 7 = 7 TrueTrue • Since (3, −4/3) checks in both equations, it is the solution • Graph confirms

  31. 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. 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. 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. 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. 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. 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. 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. Example  Solve • The Algebra Mult Both Sides of Eqn-3 by 3 Mult Both Sides of Eqn-4 by −2 Add Eqns • Solve for x using y = −2 in Eqn-3 • Students to Verify that solution (−2, −2) checks

  39. Solving Systems by Elimination • Write the equations in standard form (Ax + By = C). • Use the multiplication principle to clear fractions or decimals. • Multiply one or both equations by a number (or numbers) so that they have a pair of terms that are additive inverses. • Add the equations. The result should be an equation in terms of one variable. • Solve the equation from step 4 for the value of that variable. • 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. • Check your solution in the original equations

  40. Types of Systems of Equations • When solving problems concerning systems of two linear equations and two variables there are three possible outcomes. • Consistent Systems • INConsistent Systems • Dependent Systems

  41. Case-1 Consistent Systems • In this case, the graphs of the two lines intersect at exactly one point.

  42. 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. 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. Testing for System Case • Given System of Two Eqns • Case-1 Consistent → m1≠m2 • Case-2 Inconsistent → • m1= m2 • b1≠b2 • Case-3 Dependent → • m1= m2 AND b1=b2; • Or for a constant K: m1= Km2 AND b1= Kb2

  45. Example  Solve by Graphing • Solve System: y = 3x/4 + 2 • 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  3 • Thus the system is INCONSISTENT and has NO solution.

  46. Example  Solve System by Sub • SolveSystem • 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. 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. Solve System: Example  Solve By Graphing • SOLUTION: graph Both Equations • Both equations represent the SAME line. • Because the equations are EQUIVALENT, any solution of ONE • equation is a solution of the OTHER equation as well.

  49. Example  Solve By Subbing • SolveSystem • 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. Example  Solve By Subbing • Solve by Algebra Equation (1) Use Eqn-2 to Substitute for y Simplifying ReArranging Final ReArrangement

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