Systems of Linear Equations by 0fKplDoV

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									   5.6           Daily Warm-Up 7.1

Is the Ordered Pair a Solution of the equation 2x – 3y = 5?
1. (1, 0)
2. (-1, 1)
3. (1, -1)
4. (4, 1)
   7.1 Solving Linear Systems by Graphing

Objective: Solve a system of linear equations by
graphing
 What is a System of Linear Equations?

A system of linear equations is simply two or more linear equations
using the same variables.

We will only be dealing with systems of two equations using two
variables, x and y.

If the system of linear equations is going to have a solution, then
the solution will be an ordered pair (x , y) where x and y make
both equations true at the same time.

We will be working with the graphs of linear systems and how to find
their solutions graphically.
  How to Use Graphs to Solve Linear Systems
                                                           y
Consider the following system:
              x – y = –1
                x + 2y = 5
Using the graph to the right, we can
see that any of these ordered pairs will
make the first equation true since they                                        x
lie on the line.                                               (1 , 2)

We can also see that any of these
points will make the second equation
true.
However, there is ONE coordinate that
makes both true at the same time…
      The point where they intersect makes both equations true at the same time.
  How to Use Graphs to Solve Linear Systems
                                                        y
Consider the following system:
              x – y = –1
                x + 2y = 5
We must ALWAYS verify that your
coordinates actually satisfy both
equations.                                                                    x
                                                            (1 , 2)
To do this, we substitute the
coordinate (1 , 2) into both
equations.
   x – y = –1            x + 2y = 5
(1) – (2) = –1       (1) + 2(2) =    Since (1 , 2) makes both equations
                             1+4=5   true, then (1 , 2) is the solution to the
                                      system of linear equations.
           Graphing to Solve a Linear System
Solve the following system by graphing:
              3x + 6y = 15                Start with 3x + 6y = 15

             –2x + 3y = –3                Subtracting 3x from both sides yields
                                                      6y = –3x + 15
   While there are many different         Dividing everything by 6 gives us…
   ways to graph these equations, we
                                                             1        5
   will be using the slope – intercept                y= -   2   x+   2

   form.
                                           Similarly, we can add 2x to both
                                           sides and then divide everything by
   To put the equations in slope           3 in the second equation to get
   intercept form, we must solve both                   y=   2
                                                                 x- 1
                                                             3
   equations for y.

                      Now, we must graph these two equations
           Graphing to Solve a Linear System
Solve the following system by graphing:                          y

               3x + 6y = 15
              –2x + 3y = –3

Using the slope intercept forms of these
equations, we can graph them carefully
                                                                                   x
on graph paper.                                                          (3 , 1)
             y = - 1 x+ 5
                    2     2
                   2
              y=   3   x- 1

                                                                       Label the
Start at the y – intercept, then use the slope.
                                                                       solution!
Lastly, we need to verify our solution is correct, by substituting (3 , 1).
Since 3(3)+ 6(1)= 15 and - 2(3)+ 3(1)= - 3, then our solution is correct!
          Graphing to Solve a Linear System
  Let's summarize! There are 4 steps to solving a linear system using a graph.

Step 1: Put both equations in            Solve both equations for y, so that
slope – intercept form                   each equation looks like
                                                      y = mx + b.
Step 2: Graph both equations             Use the slope and y – intercept for
on the same coordinate plane             each equation in step 1. Be sure to
                                         use a ruler and graph paper!

Step 3: Estimate where the               This is the solution! LABEL the
graphs intersect.                        solution!


Step 4: Check to make sure your          Substitute the x and y values into both
solution makes both equations true.      equations to verify the point is a
                                         solution to both equations.
          Graphing to Solve a Linear System
 Let's do ONE more…Solve the following system of equations by graphing.
             2x + 2y = 3                                 y

              x – 4y = –1
                                                       LABEL the solution!
Step 1: Put both equations in slope –
intercept form                                               (1 , 1 )
                                                                  2
              y = - x+ 32
                  1        1
             y=   4   x+   4                                                 x
Step 2: Graph both equations on the
same coordinate plane
Step 3: Estimate where the graphs
intersect. LABEL the solution!

                                               2(1)+ 2(1 )= 2 + 1 = 3
Step 4: Check to make sure your                        2

solution makes both equations true.               1- 4 ( 1 ) = 1- 2 = - 1
                                                         2
          Graphing to Solve a Linear System
 It’s your turn! … Solve the following system of equations by graphing.
             3x + 2y = 2                                      y

               x = –1
                                                         LABEL the solution!
Step 1: Put both equations in slope –
intercept form                                                    (-1, 2 )
                                                                       5
                    3
              y = - 2 x+ 1
            x= - 1                                                             x
Step 2: Graph both equations on the
same coordinate plane
Step 3: Estimate where the graphs
intersect. LABEL the solution!

                                                3(-1)+ 2( 2 ) = - 3 + 5 = 2
                                                          5
Step 4: Check to make sure your
solution makes both equations true.
                                                   -1 = - 1
Summary
1. When you are asked to solve the linear equation system by graphing,
   you only need to follow the four-step:

Step 1: Put both equations in slope – intercept form:    y = mx + b

Step 2: Graph both equations on the same coordinate plane: use the y-
   intercept as the first point and then use slope to create the second
   point.

Step 3: Estimate where the graphs intersect, and also label the
   intersection which is the solution of the linear equation system.

Step 4: Check to make sure your solution makes both equations true
   because the graphing may contain some error. Substitute the x and y
   values into both equations to verify the point is a solution to both
   equations.

2. Don’t forget the special line!!!
      Guided Practice – L 7.1 DHQ

PAGE: 401 in Textbook

Numbers: 8-10 on graph paper!
* ALL ANSWERS ARE ORDERED PAIRS!
Tonight’s Homework

P. WORKSHEET 7.1
#’s 1-10 ALL
   5.6           Daily Warm-Up 7.2

Is (-3, 2) a solution of the equation?
1. 3x – y = 4                            2. x + 2y = 1


3. Is (4,0) a solution of the system of equations?
7.2 Solving Systems of Equations
    using Substitution
Objective: Use substitution method to
solve a linear system
Solving Systems of Equations
using Substitution
Steps:
     1. Solve one equation for one variable (y = ; x = ; a =)
     2. Substitute the expression from step one into the
        other equation.
     3. Simplify and solve the equation.
     4. Substitute back into either original equation to find
        the value of the other variable.
     5. Check the solution in both equations of the system.
 Example #1:                        y = 4x
                                   3x + y = -21
Step 1: Solve one equation for one variable.
            y = 4x   (This equation is already solved for y.)

Step 2: Substitute the expression from step one into
         the other equation.
            3x + y = -21
            3x + 4x = -21
Step 3: Simplify and solve the equation.
                 7x = -21
                   x = -3
                              y = 4x
                              3x + y = -21
Step 4: Substitute back into either original
         equation to find the value of the other
         variable.
         y = 4x
                             This substitution is easier
         y = 4(-3) = -12
or
          3x + y = -21
          3(-3) + y = -21
             -9 + y = -21
                  y = -12

     Solution to the system is (-3, -12).
                           y = 4x
                           3x + y = -21
Step 5: Check the solution in both equations.
          Solution to the system is (-3,-12).

      y = 4x                3x + y = -21

    -12 = 4(-3)          3(-3) + (-12) = -21

    -12 = -12                -9 + (-12) = -21
                                    -21= -21
Example #2:                  x + y = 10
                            5x – y = 2
Step 1: Solve one equation for one variable.
           x + y = 10
               y = –x +10
Step 2: Substitute the expression from step one into
        the other equation.
            5x – y = 2
            5x – (–x +10) = 2
                             x + y = 10
                            5x – y = 2
Step 3: Simplify and solve the equation.
              5x – (–x + 10) = 2
                 5x + x – 10 = 2
                    6x – 10 = 2
                        6x = 12
                          x=2
                             x + y = 10
                            5x – y = 2
Step 4: Substitute back into either original
         equation to find the value of the other
         variable.
           x + y = 10
           2 + y = 10
                y=8
        Solution to the system is (2,8).
                           x + y = 10
                           5x – y = 2
Step 5: Check the solution in both equations.

          Solution to the system is (2, 8).
    x + y =10              5x – y = 2
    2 + 8 =10             5(2) - (8) = 2
     10 =10                10 – 8 = 2
                              2=2
You try Exercise 1

Solve the system using the substitution
 method.
 y = –63 – 7x
 3x + 7y = –73

A)   (8, –8)
B)   (–8, –7)
C)   (–7, –8)
D)   (–7, 9)
YOU TRY: Solve by substitution:

   2a  3b  7
   2a  b  5
Summary
1. When solving the linear systems, it is very
  important to remember the 5 steps and follow
  them in the order.
2. When in a certain step the equation reduces to
   a contradiction, you could conclude that the
   system is inconsistent and has no solution.
3. When in a certain step the equation reduces to
   an identity, you could conclude that the system
   is dependent and has many solutions.
   Guided Practice – L 7.2 DHQ

PAGE: 408 in Textbook

Numbers: 7-8
• ALL ANSWERS ARE ORDERED
  PAIRS!
• MUST SHOW WORK
Tonight’s Homework

P. 408-410
#’s 17-24, 54-59
5.6         Daily Warm-Up 7.3

Give the OPPOSITE of each number.
1. 3              2. 1                3. -1/2


Simplify.
4.   3(2x – 7y)          5.   -2(5x + y)
   7.3 Solving Linear Systems by
 Linear Combinations (Elimination)
              Method

Objective: Use linear combinations to solve a system of
linear equations.
What is a System?
 A system of linear equations is:

 a. A set of parabolas
 b. A set of two or more lines
 c. A stereo component
Linear Combinations (Elimination) Method


 Linear Combinations (Elimination)
 method is used when it appears easy to
 eliminate one variable from the system
 through transformation. Remember that
 linear transformations do not change the
 solutions of a system.
Linear Combinations (Elimination) Method

Step 1 Make sure the variables are lined up properly
  by their names
             -2x + y = 4
             -6x + y = 0
Step 2 Make the coefficients of one of variables
  opposites (Multiply one or both equations by
  appropriate numbers so that the coefficients in one
  variable are opposites). Correctly applying the
  distributive property is very important!!!
   In this question, notice that the y coefficients are 1,
  therefore we can multiply either equation by -1 and
  add the system, thus eliminating the y variable.
Linear Combination (Elimination) Method

Step 2 Let’s transform the second equation:
           -1(-6x + y = 0)
               6x – y = 0
Step 3 Add the new two equations up:
               -2x + y = 4
            + 6x – y = 0
                4x      =4
Linear Combination (Elimination) Method

Step 4 Solving the resulted one variable equation
  for x
                       4x = 4
yields x = 1.

Step 5 Back substitute to one of the equation in the
  system to find the value for the other variable.
  Once we have the x value, we can plug it into
  either of our original equations and solve for y:
                   -2x + y = 4
                   -6x + y = 0
Linear Combination (Elimination) Method

Step 5 Plugging x = 1 into the first equation yields:
                    -2 (1) + y = 4
                    -2 + y = 4,
                    y=6

Step 6 Check the answer.
             -2(1) + 6 = 4
             -6(1) + 6 = 0

So our solution is (1, 6).
Linear Combination (Elimination) Method

Now you try
Use the elimination method to solve the system:
              x – 5y = –2
             3x + 2y = 11
The first step is:
a. Add the equations together
b. Transform the first equation by multiplying it
   by –3
c. I’m not sure. I need to review the elimination
   method.
Linear Combination (Elimination) Method


     Yes, notice that if we multiply the first
     equation by -3, we obtain additive inverses
     for the x coefficient. Our system is now:
              -3x + 15y = 6
               3x + 2y = 11
     The next step is:
a.   Add the two equations
b.   Solve the second equation for y
c.   I’m not sure. I need to review.
Linear Combination (Elimination) Method


Yes, by adding the two equations we get:
            – 3x + 15y = 6
         + 3x + 2y = 11
                     17y = 17

Solving for y, we get y = 1. The next step is:
a. Plug y = 1 into either equation and solve for
    x
b. Plug x = 1 into either equation and solve for
    y
c. I’m not sure. I need to review.
Linear Combination (Elimination) Method


 We found that y=1, not x=1. So we must plug
 y=1 into either equation to solve for x not for y.
Linear Combination (Elimination) Method

  Yes, now we can plug y = 1 into either of the
  original equations, or the transformed equation.
  Let’s choose the first original equation:
                 x – 5y = –2
                 x – 5(1) = –2
                 x – 5 = –2
                 x=3

  So our solution is (3, 1), which must be the only
  solution to a system of two lines.
  Next, we must check the solved order pair is the
  solution to the system. (omitted here)
You Try
Solve the system using elimination method:
                 2x + 5y = 7
                 3x + y = -9
The solution is:
a. (3, -4)
b. (-4, -3)
c. (-4, 3)
d. (-3, -4)
Linear Combination (Elimination) Method

Example
Use any method to solve the system:
              3g – 24 = –4h
             –2 + 2h = g
The first step is(are):
a. Add the equations together
b. Use the subsitution method, replace the first
   variable g by the expression in the second.
c. Line up the variables.
d. I’m not sure. I need to review the both
   methods.
Any Method (Substitution)

Use any method to solve the system:
               3g – 24 = –4h
              –2 + 2h = g
Plug the second variable g into the g in the first
    equation:
    3(-2 + 2h) – 24 = -4h
    -6 + 6h – 24 = -4h
    6h – 30 = -4h
    10h – 30 = 0
    10h = 30
    h=3
Go to the second equation to find out g:
    g = -2 + 2(3) = -2 + 6 = 4
The solution is (4, 3).
Any Method
(Linear Combination/Elimination)
Use the any method to solve the system:
               3g – 24 = –4h
              –2 + 2h = g
Line up the variable:
       3g + 4h = 24
         g – 2h = -2
Multiply 2 on both sides of the second equation and
    add onto the first equation:
       2g – 4h = -4
    + 3g + 4h = 24
        5g       = 20              g=4
Plug into second equation to find h:
    4 – 2h = -2
    -2h = -6         h=3
The solution is (4, 3)
Summary
1. The key concepts of the linear combinations
    (elimination) method are to
   (a) variables are lined up properly
   (b) find out if there are any opposite coefficients
       for the same variable. If yes, directly add two
       equations up and eliminate one variable. If
       not,
   (c) multiply (or divide) some number(s) to
       equation(s) so that two opposite coefficients
       for the same variable show up. Then follow
       (b).
2. Some problems can be applied both substitution
    and linear combination (elimination) methods.
    However, it is better to use linear combination
    (elimination) method than to use the substitution
    method to some other problems. Choosing an
    appropriate method is important.
   Guided Practice – L 7.3 DHQ

PAGE: 414 in Textbook

Numbers: 4-5
• ALL ANSWERS ARE ORDERED
  PAIRS!
• MUST SHOW WORK
Tonight’s Homework

P. 414
#’s 8-13, 17-19 +
QUIZ REVIEW
 WORKSHEET
        Daily Warm-Up 7.5

Rewrite the equations to Slope-Intercept Form ( y = mx + b )
 1. 5( x  1)  y  3             2.    6 x  3 y  9
Graph the equations
3. 5( x  1)  y  3              4.    6 x  3 y  9
7.5 Special Types of Liner Systems

Objectives

1) Identify linear system algebraically as
having one solution, no solution, and infinite
many solutions.

                        Slope-Intercept Form



2) Identify linear system geometrically as
having one solution, no solution, and infinite
many solutions.
               Special Types of Linear Systems

Example 1 Solve the linear equation system
                      3x  2 y  5     (1)
                       x  2y  7      (2)
Solution 1
Substitution                          3(3)  2(2)  5        (1)
   From (2)       x = –2y + 7          3  2(2)  7        (2)
   To (1)      3(–2y + 7) – 2y = 5
                                       The solution is (3, 2).
               –6y + 21 – 2y = 5
                  –8y = –16
                   y=2
   then
            x = –2(2) + 7 = 3
             Special Types of Linear Systems

Example 1 Solve the linear equation system
                     3x  2 y  5     (1)
                      x  2y  7      (2)
                                                  rise 3
Solution 2                                     m     
Graphing                                          run 2
3x  2 y  5       2 y  3x  5
    3       5           3     5
y       x         y  x
    2       2          2     2
 x  2y  7         2 y  x  7                  (3 , 2)

     1      7           1      7
 y      x         y       x
      2      2            2      2
                           rise  1
                     m         
                           run 2
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y


                                  Lines intersect
                                    one solution



                                 x
          A Linear System with No Solution

Show that this linear                 2x  y  5       Equation 1
METHOD 1: SUBSTITUTION
system has no solution.               2x  y  1       Equation 2


                                      Because Equation 2 can be revised
                                      to y  –2 x  1, you can substitute
                                      –2 x  1 for y in Equation 1.

              2x  y  5              Write Equation 1.

     2 x  (–2 x  1)  5             Substitute –2 x  1 for y.

                    15               Simplify. False statement!


     Once the variables are eliminated, the statement is not true regardless
     of the values of x and y. This tells you the system has no solution.
             A Linear System with No Solution


Show that this linear system          2x  y  5                Equation 1
METHOD 2: GRAPHING
has no solution.                      2x  y  1                Equation 2

Rewrite each equation                 y  –2 x  5              Revised Equation 1
in slope-intercept form.              y  –2 x  1              Revised Equation 2


Graph the linear system.                                6

                                                        5
                                                                         y  2x  5
                                                        4

The lines are parallel; they                            3
have the same slope but                                 2
different y-intercepts. Parallel       y  2x  1      1
lines never intersect, so the
system has no solution.                5   4    3   2   1        0   1   2   3   4   5
                                                            1
        Special Types of Linear Systems

• When solving a system of linear equations, if
  the resulting equation in certain step reduces to
  an inconsistent equation (contradiction), then
  the system has no solution and is inconsistent.
• Geometrically speaking, the two equations in
  the system represent two parallel lines which
  never intersect.
IDENTIFYING THE NUMBER OF SOLUTIONS
 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y


                                      Lines are parallel
                                         no solution



                                  x
Practice 2
Use the following method to solve the equation system

(1) Substitution
(2) Graphing

  2x – 4y = -2
  x – 2y = 0
               Special Types of Linear Systems

Practice 2 Solve the linear equation system
                      2 x  4 y  2         (1)
                       x  2y  0            (2)
Solution 1
Substitution
   From (2)         x = 2y
   To (1)        2(2y) – 4y = –2
                  4y – 4y = –2
                    0 = –2
        This is an inconsistent equation
            So the system has no solution.
               Special Types of Linear Systems

 Practice 2 Solve the linear equation system
                       2 x  4 y  2   (1)
                       x  2y  0       (2)
                                                    rise 1
 Solution 2                                      m     
                                                    run 2
 Graphing
2 x  4 y  2      4 y  2 x  2
       2      2         1   1
  y      x          y  x
       4     4         2   2
  x  2y  0         2 y   x  0
    1    0              1
 y    x             y  x0
    2    2             2
                           rise 1
                       m      
                           run 2
            A Linear System with Many Solutions


METHOD 1:this linear system
Show that LINEAR COMBINATIONS                  –2x  y  3       Equation 1
has infinitely many solutions.                – 4 x  2y  6     Equation 2

You can multiply Equation 1 by –2.



                 4x – 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 Many Solutions


Show that this linear system
METHOD 2: GRAPHING
                                     –2x  y  3        Equation 1
has infinitely many solutions.      – 4 x  2y  6      Equation 2


Rewrite each equation                  y  2x  3       Revised Equation 1
in slope-intercept form.               y  2x  3       Revised Equation 2


Graph the linear system.                                6

                                                        5                   –2x  y  3
                                       – 4x  2y  6    4
From these graphs you                                   3
can see that the equations
                                                        2
represent the same line.
Any point on the line is                                1
a solution.
                                        5   4   3   2   1       0   1   2     3   4   5
                                                            1
          Special Types of Linear Systems

• When solving a system of linear equations, if the
  resulting equation in certain step reduces to an
  identity equation, then the system has infinite
  many solutions and is dependent.
• Geometrically speaking, the two equations in the
  system represent two identical lines which
  always intersect. Any point on the first line (the
  solution to the first equation) is also on the
  second line (the solution to the second equation).
  The two identical lines have infinitely many
  intersection.
IDENTIFYING THE NUMBER OF SOLUTIONS
 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y
                                      Lines coincide
                                      infinitely many
                                          solutions
                                      (the coordinates
                                      of every point on
                                      the line)
                                  x
Practice 3
Use the following method to solve the equation system

(1) Linear Combination
(2) Graphing

   3x – 4y = -4
  -6x + 8y = 8
           Special Types of Linear Systems

Practice 3 Solve the linear equation system
                      3x  4 y  4     (1)
                      6 x  8 y  8    (2)
Solution 1
Linear Combination
Multiply 2 to (1)
       2(3x – 4y)= 2(– 4)
       6x – 8y = –8         (1)
      -6x + 8y = 8          (2)
          0=0           This is an identity. So
                        the equation system has
                        many solutions.
                 Special Types of Linear Systems

Practice 3 Solve the linear equation system
                         3x  4 y  4    (1)
                         6 x  8 y  8   (2)
Solution 2                                            rise 3
                                                   m     
Graphing                                              run 4
3x  4 y  4         4 y  3x  4
   3       4              3
y     x               y  x 1
   4      4              4
6 x  8 y  8          8 y  6x  8
     6   8                 3
  y  x                y  x 1
     8   8                 4
                             rise 3
                         m      
                             run 4
IDENTIFYING THE NUMBER OF SOLUTIONS
 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y


                                      Lines intersect
                                        one solution



                                  x
IDENTIFYING THE NUMBER OF SOLUTIONS
 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y


                                      Lines are parallel
                                         no solution



                                  x
IDENTIFYING THE NUMBER OF SOLUTIONS
 NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
                     y
                                      Lines coincide
                                      infinitely many
                                          solutions
                                      (the coordinates
                                      of every point on
                                      the line)
                                  x
      IDENTIFYING THE NUMBER OF SOLUTIONS
CONCEPT        NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
SUMMARY



           y                        y                           y




                    x                         x                               x




  Lines intersect        Lines are parallel          Lines coincide
    one solution             no solution          infinitely many solutions

  Independent              Inconsistent               Dependent
Summary

Similar to the situations when solving the one variable
  linear equations, solving the linear systems has the
  similar 3 situations:
   (a) One solution -- two lines intersect -- the
      system is independent
   (b) No solution -- two lines are parallel -- the
      system is inconsistent -- the system contains a
      contradiction
   (c) Many solution -- two lines are coincide --
      the system is dependent -- the system
      contains an identity
    Guided Practice – L 7.3 DHQ


PAGE: 429 in Textbook

Numbers: 6-8
• USE ANY METHOD
• ALL ANSWERS ARE ORDERED
  PAIRS!
 Tonight’s Homework

P. 429-430
#’s 12-21, 24-27
• MUST SHOW WORK

								
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