Chapter 3

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• Chapter 3 Linear Systems
• Systems of Linear Equations
• Solving Systems of Equations by Substitution

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Systems of Equations

• A set of equations is called a system of
  equations.
• The solutions must satisfy each equation in the
  system.
• If all equations in a system are linear, the
  system is a system of linear equations, or a
  linear system.

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• Systems of Linear Equations
• A solution to a system of equations is an ordered
  pair that satisfy all the equations in the
  system.
• A system of linear equations can have
• 1. Exactly one solution
• 2. No solutions
• 3. Infinitely many solutions

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• Systems of Linear Equations
• There are four ways to solve systems of linear
  equations
• 1. By graphing
• 2. By substitution
• 3. By addition (also called elimination)
• 4. By multiplication

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• Solving Systems by Graphing
• When solving a system by graphing
• Find ordered pairs that satisfy each of the
  equations.
• Plot the ordered pairs and sketch the graphs of
  both equations on the same axis.
• The coordinates of the point or points of
  intersection of the graphs are the solution or
  solutions to the system of equations.

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• Solving Systems by Graphing

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Linear System in Two Variables

• Three possible solutions to a linear system in
  two variables
• One solution coordinates of a point
• No solutions inconsistent case
• Infinitely many solutions dependent case

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2x y 2 x y -2
2x y 2 -y -2x 2 y 2x 2
x y -2 y -x - 2
Different slope, different intercept!
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3x 2y 3 3x 2y -4
3x 2y 3 2y -3x 3 y -3/2 x 3/2
3x 2y -4 2y -3x -4 y -3/2 x - 2
Same slope, different intercept!!
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x y -3 2x 2y -6
x y -3 -y -x 3 y x 3
2x 2y -6 -2y -2x 6 y x 3
Same slope, same intercept! Same equation!!
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• Determine Without Graphing
• There is a somewhat shortened way to determine
  what type (one solution, no solutions, infinitely
  many solutions) of solution exists within a
  system.
• Notice we are not finding the solution, just what
  type of solution.
• Write the equations in slope-intercept form y
  mx b.
• (i.e., solve the equations for y, remember that
  m slope, b y - intercept).

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• Determine Without Graphing
• Once the equations are in slope-intercept form,
  compare the slopes and intercepts.
• One solution the lines will have different
  slopes.
• No solution the lines will have the same slope,
  but different intercepts.
• Infinitely many solutions the lines will have
  the same slope and the same intercept.

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• Determine Without Graphing
• Given the following lines, determine what type of
  solution exists, without graphing.
• Equation 1 3x 6y 5
• Equation 2 y (1/2)x 3
• Writing each in slope-intercept form (solve for
  y)
• Equation 1 y (1/2)x 5/6
• Equation 2 y (1/2)x 3
• Since the lines have the same slope but different
  y-intercepts, there is no solution to the system
  of equations. The lines are parallel.

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• Substitution Method
• Procedure for Substitution Method
• 1. Solve one of the equations for one of the
  variables.
• 2. Substitute the expression found in step 1 into
  the
• other equation.
• 3. Now solve for the remaining variable.
• 4. Substitute the value from step 2 into the
  equation
• written in step 1, and solve for the
  remaining variable.

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• Substitution Method
• 1. Solve the following system of equations by
  substitution.

Step 1 is already completed.
Step 2Substitute x3 into 2nd equation and solve.
Step 3 Substitute 4 into 1st equation and solve.
The answer ( -4 , -1)
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1) Solve the system using substitution

• x y 5
• y 3 x

Step 1 Solve an equation for one variable.
The second equation is already solved for y!
Step 2 Substitute
x y 5x (3 x) 5
2x 3 5 2x 2 x 1
Step 3 Solve the equation.
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1) Solve the system using substitution

• x y 5
• y 3 x

x y 5 (1) y 5 y 4
Step 4 Plug back in to find the other variable.
(1, 4) (1) (4) 5 (4) 3 (1)
Step 5 Check your solution.
The solution is (1, 4). What do you think the
answer would be if you graphed the two equations?
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2) Solve the system using substitution

• 3y x 7
• 4x 2y 0

It is easiest to solve the first equation for
x. 3y x 7 -3y -3y x -3y 7
Step 1 Solve an equation for one variable.
Step 2 Substitute
4x 2y 0 4(-3y 7) 2y 0
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2) Solve the system using substitution

• 3y x 7
• 4x 2y 0

-12y 28 2y 0 -14y 28 0 -14y -28 y 2
Step 3 Solve the equation.
4x 2y 0 4x 2(2) 0 4x 4 0 4x 4 x 1
Step 4 Plug back in to find the other variable.
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2) Solve the system using substitution

• 3y x 7
• 4x 2y 0

Step 5 Check your solution.
(1, 2) 3(2) (1) 7 4(1) 2(2) 0
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• Deciding whether an ordered pair is a solution
  of a linear system.
• The solution set of a linear system of equations
  contains all ordered pairs that satisfy all the
  equations at the same time.
• Example 1 Is the ordered pair a solution of the
  given system?
• 2x y -6 Substitute the ordered pair into
  each equation.
• x 3y 2 Both equations must be satisfied.
• A) (-4, 2) B) (3, -12)
• 2(-4) 2 -6 2(3) (-12) -6
• (-4) 3(2) 2 (3) 3(-12) 2
• -6 -6 -6 -6 2 2
  -33 ? -6
• ? Yes ? No

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Substitution Method

• Example Solve the system.
• Solution

(1) (2)
Solve (2) for y.
Substitute y x 3 in (1).
Solve for x.
Substitute x 1 in y x 3.
Solution set (1, 4)
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Systems of Linear Equations in Two Variables

• Solving Linear Systems by Graphing.
• One way to find the solution set of a linear
  system of equations is to graph each equation and
  find the point where the graphs intersect.
• Example 1 Solve the system of equations by
  graphing.
• A) x y 5 B) 2x y -5
• 2x - y 4 -x 3y 6
• Solution (3,2) Solution (-3,1)

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Systems of Linear Equations in Two Variables

• Solving Linear Systems by Graphing.
• There are three possible solutions to a system
  of linear equations in two variables that have
  been graphed
• 1) The two graphs intersect at a single point.
  The coordinates give the solution of the system.
  In this case, the solution is consistent and
  the equations are independent.
• 2) The graphs are parallel lines. (Slopes are
  equal) In this case the system is inconsistent
  and the solution set is 0 or null.
• 3) The graphs are the same line. (Slopes and
  y-intercepts are the same) In this case, the
  equations are dependent and the solution set is
  an infinite set of ordered pairs.

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4-1 Systems of Linear Equations in Two Variables

• Solving Linear Systems of two variables by
  Method of Substitution.
• Step 1 Solve one of the equations for either
  variable
• Step 2 Substitute for that variable in the
  other equation
• (The result should be an equation with just one
  variable)
• Step 3 Solve the equation from step 2
• Step 4 Substitute the result of Step 3 into
  either of the original equations and solve for
  the other value.
• Step 6 Check the solution and write the
  solution set.

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4-1 Systems of Linear Equations in Two Variables

• Solving Linear Systems of two variables by
  Method of Substitution.
• Example 6 Solve the system 4x y 5
• 2x - 3y 13
• Step 1 Choose the variable y to solve for in
  the top equation
• y -4x 5
• Step 2 Substitute this variable into the bottom
  equation
• 2x - 3(-4x 5) 13 2x 12x -
  15 13
• Step 3 Solve the equation formed in step 2
• 14x 28 x 2
• Step 4 Substitute the result of Step 3 into
  either of the original equations and solve for
  the other value. 4(2) y 5
• y -3
• Solution Set (2,-3)
• Step 5 Check the solution and write the
  solution set.

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Systems of Linear Equations in Two Variables

• Solving Linear Systems of two variables by
  Method of Substitution.
• Example 7
• Solve the system
• y -2x 2
• -2x 5(-2x 2) 22
  -2x - 10x 10 22
• -12x 12
• x -1 2(-1) y 2
• y 4
• Solution Set (-1,4)

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Your Turn
3x y 4 x 4y - 17
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Your Turn
2x 4y 4 3x 2y 22
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Clearing Fractions or Decimals
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Systems without a Single Point Solution
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0 4 untrueInconsistent Systems - how can you
tell?

• An inconsistent system has no
  solutions.
• (parallel lines)
• Substitution Technique

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0 0 or n n Dependent Systems how can
you tell?

• A dependent system hasinfinitely many solutions.
• (its the same line!)Substitution
  Technique

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• Modeling Examples
• The reason to learn about systems of equations is
  to learn how to solve real world problems.
• Study Example 8 on page 360 in the text. Notice
  how the original equations are set up based on
  the data in the question.
• Also note that we are trying to determine when
  the total cost at each garage will be the same.
  To do this, set the two cost equations equal to
  each other and solve. You will see this type of
  problem often.

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• Modeling Examples
• Study Example 9 on page 361 in the text. This is
  a mixture problem. Notice how the original
  equations are set up based on the data in the
  question.
• Once the equations are set up, the 2nd equation
  is multiplied by 100 to remove the decimal. This
  is a common occurrence, so make sure you know how
  to do this.
• Note The example is solved using the addition
  method. It can also be solved by substitution.

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• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.

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• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.
• First assign the variables
• let x of 2 point shots
• let y of 3 point shots

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• Modeling Examples
• 4. Read problem 40 on page 362 of the text
  basketball game.
• First assign the variables
• let x of 2 point shots
• let y of 3 point shots
• Writing the 1st equation
• They made 45 goals in a recent game
• x y 45

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• Modeling Examples
• 4 continued.
• Writing the 2nd equation

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• Modeling Examples
• 4 continued.
• Writing the 2nd equation
• Some 2 pointers, some 3 pointers, for a total
  score of 101 points
• 2x 3y 101

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• Modeling Examples
• 4 continued.
• Writing the 2nd equation
• Some 2 pointers, some 3 pointers, for a total
  score of 101 points
• 2x 3y 101
• In words, the equation says 2 times the number of
  2 point shots plus 3 times the number of 3 point
  shots totals 101 points.

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• Modeling Examples
• 4 continued.
• The two equations are
• x y 45
• 2x 3y 101

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• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101

Lets eliminate x, multiply the entire 1st
equation by 2.
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• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101
• -2x -2y -90
• 2x 3y 101

Lets eliminate x, multiply the entire 1st
equation by 2.
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• Modeling Examples
• 4 continued.
• The two equations are
• -2( x y 45 )
• 2x 3y 101
• -2x -2y -90
• 2x 3y 101
• y 11

• Add down to eliminate x. Substitute y into the
  1st equation. x 11 45, so x 34.
• - 2 point shots and 11 - 3 point shots.
  

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• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.

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• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.
• First assign the variables
• let x gallons of 5 milk
• let y gallons of skim (0) milk

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• Modeling Examples
• 5. Read problem 44 on page 363 in the text
• A Milk Mixture.
• First assign the variables
• let x gallons of 5 milk
• let y gallons of skim (0) milk
• Writing the 1st equation
• x y 100
• This is because they want to make a mixture
  totaling 100 gallons of milk.

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• Modeling Examples
• 5. Continued
• Writing the 2nd equation

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• Modeling Examples
• 5. Continued
• Writing the 2nd equation
• 0.05x 0.0y 0.035(100)
• Basically, we are multiplying the 1st equation by
  the percent butterfat of the milk. Our final
  mixture should be 3.5, so we multiply
  0.035(100), since we want 100 total gallons.

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• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)

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• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)
• Next, multiply the 2nd equation by 1000 to remove
  the decimal. This gives us the following system
  of equations x y 100
• 50x 0y 35(100)

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• Modeling Examples
• 5. Continued
• The two equations are
• x y 100
• 0.05x 0.0y 0.035(100)
• Next, multiply the 2nd equation by 1000 to remove
  the decimal. This gives us the following system
  of equations x y 100
• 50x 0y 35(100)
• Solve the system (use substitution since the 2nd
  equation has only one variable). The answer
  follows on the next slide.

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• Modeling Examples
• 5. Continued
• The answer is 70 gallons of 5 milk and 30
  gallons of skim (0) milk.

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• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.

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• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.
• First assign the variables
• let x of adult tickets sold (5 per ticket)
• let y of student tickets sold (2 per ticket)

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• Modeling Examples
• 6. Read problem 48 on page 363 in the text
  School Play Tickets.
• First assign the variables
• let x of adult tickets sold (5 per ticket)
• let y of student tickets sold (2 per ticket)
• Writing the 1st equation
• x y 250
• Since a total of 250 tickets were sold.

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• Modeling Examples
• 6. Continued
• Writing the 2nd equation

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• Modeling Examples
• 6. Continued
• Writing the 2nd equation
• 5x 2y 950
• Basically, we multiplied the 1st equation by the
  price of the tickets, and set it equal to the
  amount of money collected.

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• Modeling Examples
• 6. Continued
• Writing the 2nd equation
• 5x 2y 950
• Basically, we multiplied the 1st equation by the
  price of the tickets, and set it equal to the
  amount of money collected.
• Do you see how this is similar to example 4?
  The 2 and 3 point shots?

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• Modeling Examples
• 6. Continued
• The two equations are
• x y 250
• 5x 2y 950

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• Modeling Examples
• 6. Continued
• The two equations are
• x y 250
• 5x 2y 950
• Can you solve the system using either
  substitution or addition? The answer follows on
  the next slide.

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• Modeling Examples
• 6. Continued
• The answer is 150 adult tickets were sold, and
  100 student tickets were sold.

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• Congratulations!
• You have finished the PowerPoint slides for
  Chapter 7!

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