KSU M10005 Chapter 4 Systems of Linear Equations and

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					KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

     4.1 Solving SYSTEMS of Linear Equations by                                                         y

                      GRAPHING                                                                  10



4.1.1 Determine if an ordered pair is a solution of a
 system of equations in two variables                                                              5




    A System of Linear Equations consists of 2 or more linear             -10         -5                            5        10
                                                                                                                             x
     equations.
                                                                                                   -5


    A Solution of this system consists of a point(s) that are
     common to all linear equations. We will only study                                         -10

     linear equations in two variables.
                                                                         The 'solution' to the system of equations
    The solution to these systems consists of a common                   is the common point of intersection
     intersection point. This intersection point is an
     ordered pair that is common to both equations.

                                                                1                                                                        2




    (4.1.1)To determine if a given ordered pair                     4.1.1 Examples: Which of the ordered pairs are solutions?
     is a solution:
                                                                                           2x            y 5            Order Pairs:
    Substitute the ordered pair into both                                  (a) System:                                    (i) (5,0)
                                                                                           x            3y 5
     equations and determine if it is a solution                                                                          (ii) (1,2)
     to both equations.                                                                                                   (ii) (2,1)
                                                                                              3x             y 5
                                                                                (b) System:
                                                                                              x             2y 11       Order Pairs:
                                                                                                                          (i) (2-1)
                                                                                                                          (ii) (3,4)
                                                                                                                          (ii) (0, -5)



                                                                3                                                                        4




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

4.1.2 Solve a system of linear equations by graphing
                                                           'Solution Point' for the system:
     Since a           of a system of two linear              y = -2x + 6
    equations is a common ordered pair, or point,             y = 3x - 4
    this will be the ordered pair of the point of                                                10
                                                                                                        y


    intersection of the graphs.
                                                                                                    5

     By drawing the two graphs, the          can be
            .
                                                                  -10            -5                                    5             10
                                                                                                                                     x



                                                                                                  -5




                                                                                                 -10




                                                       5                                                                                               6




 (4.1.2)

 A system that has at LEAST one intersection point is
defined as a consistent system.

 A system with NO intersection point is an nconsistent
                                                                        Line 1             Line 2           Solution       consistency    dependency
system.                                                                                                      Point
                                                           a.   y = 3x - 4            y=x+2
Systems with                graphs are termed              b.   2x + y=0              3x + y=1
                                                           c.   2x + y = 4            x+y=2
                                                           d.   x - 2y =2             3x + 2y = -2
  Systems with IDENTICAL, overlapping           are
termed
                                                           (Compare your solution graphs to the following slides)
                                                       7                                                                                               8




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

 (4.1.2 solutions for examples (a) and (b)                           (4.1.2 solutions for examples (c) and
                                                                     (d)
                               y
                         10                                                                                    y
                                                                                                         10


                                                                                                                   ( c)
                          5
                                        (a)                                                                               2x + y = 4
                                              y = 3x – 4
                                              y=x +2
                                                                                                          5
                                                                                                                          x+y=2

        -10      -5                 5                      10
                                                           x                 -10               -5                              5       10
                                                                                                                                       x


                          -5
                                        (b)                                                               -5

                                              2x + y = 0                            (d)
                                              3x + y = 1                                  x – 2y = 2
                         -10
                                                                                          3x + 2y = -2   -10




                                                                 9                                                                          10




4.1.3 Without graphing, you can also determine the                   (4.1.3)There are also methods of determining
 number of solutions of a system                                                    a system will have.

    Note:                                                                Write the two equations in slope-intercept
     Graphing is not an accurate method of                               form: y = mx + b
    determining the point of intersection.
                                                                              If both have the same slope and different
     There are other methods of accurately                                    intercepts, then the system has NO
    determining the intersection point.                                       SOLUTION and is INCONSISTENT.
                                                                             The two lines are Parallel!


                                                                                   If the two equations have different slopes,
                                                                                   then they must intersect and therefore has
                                                                11
                                                                                   ONE SOLUTION and is CONSISTENT 12




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

                                                                          The 'solution point' can also be determined,
                                                                          more accurately, by using one of two 'algebraic
                                                                          methods'.

              Line 1        Line 2      Orientation   Number of           The two algebraic methods are:
                                                      Solutions
      a.    3x + y = 1    3x + 2y = 6
                                                                              4.2: The 'substitution method'
      b.    2x + y = 0    2y = 6 - 4x
      c.    3y - 2x = 6   x + 2y = 9
                                                                              4.3: The 'addition method'
      d.    8y + 6x = 4   4y - 2 = 3x

                                                                          Both methods use algebraic methods using two
                                                                          equations to find the common point (x,y) that
                                                                          they have in common.
                                                                  13
                                                                          This point is their intersection point.           14




   4.2 Solving Systems of Linear Equations by the                      (4.2.1) Examples: Solve each of the following system of
                                                                       equations using the substitution method
4.2.1 To determine the solution point:                                    a.    x    y 20                     Solution Point:
           Solve one of the equations for either 'x' or                         x 3y
          'y' (which ever seems faster)
                                                                               y 3x 1
           Substitute this expression for the                            b.
                                                                               4y  8x 12
           corresponding 'x' or 'y' variable in the
           other equation.
                                                                         c.    x    2y 6
           Solve for the other variable                                        2x    3y 8
           Substitute this value of the other variable                         x    2y 10
           into either of the original equations to find                 d.
                                                                  15           2x    3y 18                                  16
           the value for the remaining variable.




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

                                                           (4.3.1)
   4.3 Solving Systems of Linear Equations by the                    Multiply one of the equations by a positive
               ADDITION METHOD                                       or negative number which will result in
                                                                     eliminating one of the variables if the two
                                                                     equations are added.
4.3.1Another method of determining the solution                         (One equation's coefficient of 'x' or 'y'
point is by the addition or 'elimination' method.                         will be the negative of that
                                                                          in the other equation)
This is based on the addition property of equality:                  Find the value for the remaining variable.
         If A = B and S = T then A + S = B + T                       Substitute this back into one of the original
         Also:                                                       equations to find the other coordinate.
               If A = B
                  S=T
               A+S=B+T                                17                                                            18




                                                            4.4 Systems of Linear equations and Problem Solving

                                       x   y
  a.    4x    y   13                          1                  Write each true statement as an equation in
                                d.     3   6
        2x    y   5                                             variables which define the unknowns.
                                       x   y
                                             0
  b.    2x    y 5                      2   4
                                                                 Use any of the system solution processes to find
        4x    2y 12                                             the solution for this system.
                                       x 5    y 14
                                e.       2      4
  c.    4x    2y 2                                         Note: the Substitution or Addition Method are the
        3x    2y 12                    x   2y     2
                                                           most accurate methods of determining the intersection
                                       3      6            point, while graphing is the least accurate method.


                                                      19                                                            20




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

(4.4.1 )Examples: Without solving each problem, choose each
correct solution by deciding which choice satisfies the given
conditions by substituting the choices into the equations. Which
choice will make the equation true.
                                                                                           4.   Two numbers total 83 and have a difference of 17.
      1       Two CDs and 4 cassettes cost a total of $40. However, 3 CD's                      Find the numbers
              and 5 cassettes cost $55. Find the price of each.
           a.   CD = $12; cassette = $4
                                                                                           5.   One number is 4 more than twice the second. Their
           b.   CD = $15; cassette = $2
                                                                                                total is 25. Find the numbers.
           c.   CD = $10; cassette = $5

      2       A chemistry lab stores 28 gallons of saline solution in two                  6.   In investigating train fares from Cleveburg to NYC, a
              containers. One container holds three times the capacity of                       customer finds that 3 adults and 4 children must pay
              the other. Find the capacity of each container.                                   $159, while 2 adults and 3 children must pay $112.
           a.    15 gallons: 5 gallons                                                          Find the price of the adult and child's tickets.
           b.    20 gallons: 8 gallons
           c.    21 gallons: 7 gallons

                                                                             21                                                                     22




                                                                                  (4.5.1)Inequalities have 'half planes' sets of points that
            4.5 Graphing Linear INEQUALITIES                                      satisfy the original linear inequality:

4.5.1 Graphing a linear inequality in two variables                                 y     -1.5x + 5

       The solution for                     include all points                    Use 'Test Points' to find the correct half plane.
      the line.

       The 'solution' for                        all points                             TP1: (0,2)
                or        the line, depending on the type                               TP2: (5,0)
      of inequality.
         This results in a 'half-plane' of solution points.

Terms: half-planes, boundary
                                                                             23                                                                     24




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

 Graph the inequality as if it were a linear equation to find the
boundary line. (i.e. y = mx+b)                                         4.5.1 Examples:

    Use a dotted boundary line if the inequality is < or >             Determine which of the ordered pairs are solutions
    Use a solid boundary line if the inequality contains the =         of the linear inequality in two variables.
    sign ( , or ).
                                                                       1. y – x < -2                                          (2, 1) or (5, -1)
 Choose a test point (TP), that is not on the line. Substitute the
coordinates of the TP back into the original inequality.
                                                                       2. 3x – 5y <= -4                                                (-1, 4) or (4, 0)
 If the substitution results in a true condition, shade this side of
the boundary line, otherwise shade the other side of the
boundary line.

 You may also use arrows on the boundary line ends, in place
of shading, to denote the correct half-plane of solution points

                                                                 25                                                                                                               26




4.5.1 Examples: Graph the following inequalities:                                                10y                                                   10y

Shade the correct the half-plane that is the 'solution'.                                                                                    5. 6x – 2y 0
                                                                                                                                                        5
                                                                       3. 2x + y             4    5
                                                                                                                  TP=(1,0)
                                                                       -10         -5                               5        10             -10   -5                5      10
                                                                                                                             x                                             x


                   3. 2x       y         4                                                       -5                                                    -5



                                                                                                 -10                                                   -10


                   4. x       2y         3

                   5. 6x       2y            0                                                         10
                                                                                                         y
                                                                                                                                                        10y

                                                                                                             4.     x + 2y 3
                                                                                                        5
                                                                                                                                                           5
                        1          1
                   6.     x          y           1
                        2          3                                         -10        -5                              5         10
                                                                                                                                  x         -10   -5                 5      10
                                                                                                                                                                            x

                                                                                                        -5
                                                                                                             TP=(1,0)
                                                                                                                                                        -5     6. ½ x – 1/3 y    -1
                                                                                                       -10
                                                                                                                                                       -10
     (Compare your graphs to the following slide)
                                                                 27                                                                                                               28




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

                                                                          4.6.1 To find the 'solution' to a set of inequalities,
 4.6 Graphing SYSTEMS of Linear
Note:                                                                     Graph each inequality as in Section 4.5, both on the
                                                                          same graph.
Finding the 'solution' to a system of linear
EQUATIONS finds their                                                               Graph the boundary lines (dashed or solid).

                                                                                    Use a test point for each inequality and determine
Finding the 'solution' to a system of linear                                        which half-plane is the solution for each inequality
INEQUALITIES finds the                                                              Shade that side of the boundary line, or use arrows on
       that satisfy each of the two inequalities.                                   the end of each line to denote the solution half-plane
                                                                                    for each inequality.

                                                                                    The 'solution' to the 'system' is the overlapping shaded
                                                                                    regions.
                                                                     29                                                                            30




 4.6.1 Graphing 'systems' of inequalities:                                 4.6.1 Examples: Graph the solution to the
         y<x+4                                                             following systems of inequalities. Compare your
         y > -1.5x + 5                                                     solutions to the following slides.
  Use Test Points for each inequality
  Find half plane of each                                                  1. y          x     4       3.       y    x       4       5.   x   2y        6
  Determine their intersection region  10
                                             y

                                                                                y       2x     5                    1                     x   2y        4
                         y -1.5x + 5
                                                                                                            y         x          2
                                        5
                                                                                                                    2


                   -10        -5                            5   10

                                                 TP=(2,0)
                                                                x
                                                                                         2             4.   2x           y       4
                                                                           2.       y      x       5
                                        -5                                               3
                         y x+4
                                                                                                            x       y        5
                                                                                        1
                                       -10
                                                                                y         x    3
                                                                     31
                                                                                        4                                                          32




E. Millspaw
KSU M10005 Chapter 4: Systems of Linear Equations and Inequalities

                  10
                        y
                                      1.                                                                                          3.            10
                                                                                                                                                     y


                                               y -x -4                                                                                 y x–4
                   5                                                                                                                             5

                                               y 2x + 5                                                                                y ½x+2
 -10   -5                                  5                    10                                                               -10     -5              5          10
                                                                x                                                                                                   x
                                 TP
                   -5                                                                                                                           -5       TP=(6,0)

                  -10                                                                    10
                                                                                               y   2.                                           -10                                 10
                                                                                                                                                                                         y

                                                                                                        y       2/3 x – 5
                                                                         TP=(0,4)         5                                                                                          5
                                                                                                        y       ¼x-3                                                                         TP = (4,0)
                                                          -10              -5                               5          10                                     -10          -5                 5       10
                                                                                                                       x                                                                              x
                                                                                                                                                                     4.
                                                                                          -5                                                                                        -5
                                                                                                                                                                          2x + y 4
                                                                                         -10                                                                                        -10
                                                                                                                                                                          x+y   5

                                                                                                                            33                                                                             34




                                                 y
                                           10




                                            5




            -10             -5                                       5              10
                                                                                    x


                                            -5
                                                     6.
                                                                x + 2y          6
                                           -10
                                                                x – 2y 4




                                                                                                                            35




E. Millspaw