1_19_2016_Solving Systems of Linear Equations 2

7/24/2019 1_19_2016_Solving Systems of Linear Equations 2

1/64

1

inear Systems

Systems of Linear Equations

Solving Systems of Equations bySubstitution


2/64

Systems of Equations

A set of equations is called a system ofequations.

The solutions must satisfy each equation in thesystem.

If all equations in a system are linear, the systemis a system of linear equations, or a linearsystem.


3/64

3

Systems of Linear Equations:

A solution to a system of equations is anordered pair that satisfy all the equations inthe system.

A system of linear equations can have:

1. Exactly one solution

2. No solutions

3. Infinitely many solutions


4/64

4

Systems of Linear Equations:

There are four ways to solve systems of linearequations:

1. By graphing

2. By substitution

3. By addition (also called elimination)

4. By multiplication


5/64

5

Solving Systems by Graphing:

When solving a system by graphing:1. Find ordered pairs that satisfy each of the

equations.

2. Plot the ordered pairs and sketch thegraphs of both equations on the same axis.

3. The coordinates of the point or points ofintersection of the graphs are the solution or

solutions to the system of equations.


6/64

6

Solving Systems by Graphing:

Consistent DependentInconsistent

One solution

Lines intersect

No solution

Lines are parallel

Infinite number ofsolutions

Coincide-Same

line


7/64

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

Linear System in Two Variables


8/648

2xy = 2

x + y = -2

2xy = 2

-y = -2x + 2y = 2x2

x + y = -2y = -x - 2

Different slope, different intercept!


9/649

3x + 2y = 3

3x + 2y = -4

3x + 2y = 3

2y = -3x + 3y = -3/2 x + 3/2

3x + 2y = -42y = -3x -4

y = -3/2 x - 2

Same slope, different intercept!!


10/64

xy = -3

2x2y = -6

xy = -3

-y = -x3y = x + 3

2x2y = -6-2y = -2x6

y = x + 3Same slope, same intercept!

Same equation!!


11/6411

Determine Without Graphing:

There is a somewhat shortened way todetermine what type (one solution, nosolutions, infinitely many solutions) ofsolution exists within a system.

Notice we are not finding the solution, justwhat type of solution.

Write the equations in slope-intercept form:

y = mx + b.(i.e., solve the equations for y, rememberthat m = slope, b = y - intercept).


12/6412


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 thesame slope and the same intercept.


13/6413


Given the following lines, determine what typeof 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 butdifferent y-intercepts, there is no solution to thesystem of equations. The lines are parallel.


14/64

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 theother 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 remainingvariable.


15/64

Substitution Method:

1. Solve the following system of equations by

substitution.

5

3

yx

xy

5)3( xx

532 x

82 x

4x

Step 1 is already completed.

Step 2:Substitute x+3 into2nd equation and solve.

Step 3: Substitute4 into 1st

equation and solve.

134

3

y

y

xy

The answer: ( -4 , -1)


16/64

1) Solve the system using substitution

x + y = 5y = 3 + x

Step 1: Solve an

equation for one

variable.

Step 2: Substitute

The second equation is

already solved for y!

x + y = 5

x + (3 + x) = 5

Step 3: Solve the

equation.

2x + 3 = 52x = 2

x = 1


17/64


x + y = 5y = 3 + x

Step 4: Plug back in to

find the other

variable.

x + y = 5

(1) + y = 5

y = 4

Step 5: Check your

solution.

(1, 4)

(1) + (4) = 5

(4) = 3 + (1)

The solution is (1, 4). What do you think the answerwould be if you graphed the two equations?


18/64


3y + x = 7

4x 2y = 0

Step 1: Solve an

equation for one

variable.

Step 2: Substitute

It is easiest to solve the

first equation for x.

3y + x = 7

-3y -3y

x = -3y + 7

4x2y = 0

4(-3y + 7)2y = 0


19/64


3y + x = 74x 2y = 0

Step 4: Plug back in to

find the othervariable.

4x2y = 0

4x2(2) = 0

4x4 = 0

4x = 4

x = 1

Step 3: Solve theequation.

-12y + 282y = 0

-14y + 28 = 0-14y = -28

y = 2


20/64


3y + x = 74x 2y = 0

Step 5: Check your

solution.

(1, 2)

3(2) + (1) = 7

4(1)2(2) = 0


21/64

Deciding whether an ordered pair is asolution of a linear system.The solution set of a linear system of equations contains all orderedpairs 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 = -62 = 2 -33 -6

Yes No


22/64

Substitution Method

Example Solve the system.

Solution

4

31

155

11623

11)3(233

y

y

xx

xx

xxxy Solve (2) fory.

Substitutey =x + 3 in (1).

Solve forx.

Substitutex = 1 iny =x + 3.

Solution set: {(1, 4)}

3

1123

yx

yx (1)

(2)

S f Li E i i T V i bl


23/64

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 grapheach equation and find the point where the graphs intersect.

Example 1: Solve the system of equations bygraphing.

A) x + y = 5 B) 2x + y = -5

2x - y = 4 -x + 3y = 6

Solution: {(3,2)} Solution: {(-3,1)}

S t f Li E ti i T V i bl


24/64


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 isinconsistent 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.

4 1 S t f Li E ti i T V i bl


25/64

4-1 Systems of Linear Equations in Two Variables

Solving Linear Systems of two variables byMethod 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 originalequations and solve for the other value.

Step 6: Check the solution and write the solution set.

4 1 S stems of Linear Eq ations in T o Variables


26/64

4-1 Systems of Linear Equations in Two Variables


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 = 2Step 4: Substitute the result of Step 3 into either of the original equations and solve for the other

value. 4(2) + y = 5y = -3

Solution Set: {(2,-3)}Step 5: Check the solution and write the solution set.

S t f Li E ti i T V i bl


27/64



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)}

1 1 1

2 4 2

2

1 1 1 rewrite as 4[ ] 2 2

2 4 2

: 2 2

-2 5 2

5 2

2

2

x y

x y

x y x y

Solve x y

x y


28/64

3x y = 4x = 4y - 17

Your Turn:


29/64

Your Turn:

2x + 4y = 43x + 2y = 22


30/64

Clearing Fractions or Decimals


31/64

Systems without a Single Point Solution


32/64

0 = 4 untrue

Inconsistent Systems - how can you tell?

An inconsistent systemhas no solutions.

(parallel lines)

Substitution Technique

ntinconsiste

xx

xx

xyBxyA

072225 33

2353

23)(53)(


33/64

0 = 0 or n = n

Dependent Systemshow can you tell?

A dependent system hasinfinitely many solutions.

(its the same line!)Substitution Technique

dependentyy

yyAyx

yxB

xyBxyA

666633

6)3(23)(3

24128)(

24812)(623)(

23

23


34/64

34


35/64

35

Modeling Examples:

The reason to learn about systems of equationsis 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 whenthe total cost at each garage will be the same.To do this, set the two cost equations equal toeach other and solve. You will see this type ofproblem often.


36/64

36

Modeling Examples:

Study Example 9 on page 361 in the text. Thisis a mixture problem. Notice how the originalequations are set up based on the data in thequestion.

Once the equations are set up, the 2nd equationis multiplied by 100 to remove the decimal.This is a common occurrence, so make sureyou know how to do this.

Note: The example is solved using the additionmethod. It can also be solved by substitution.


37/64

37

Modeling Examples:

4. Read problem 40 on page 362 of the textbasketball game.


38/64

38

Modeling Examples:


First assign the variables:

let x = # of 2 point shots

let y = # of 3 point shots


39/64

39

Modeling Examples:



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


40/64

40

Modeling Examples:

4 continued.

Writing the 2nd equation:


41/64

41

Modeling Examples:

4 continued.


Some 2 pointers, some 3 pointers, for a totalscore of 101 points

2x + 3y = 101


42/64

42

Modeling Examples:

4 continued.


Some 2 pointers, some 3 pointers, for a totalscore of 101 points

2x + 3y = 101

In words, the equation says 2 times the numberof 2 point shots plus 3 times the number of 3

point shots totals 101 points.


43/64

43

Modeling Examples:

4 continued.

The two equations are:

x + y = 45

2x + 3y = 101


44/64

44

Modeling Examples:

4 continued.


-2( x + y = 45 )

2x + 3y = 101

Lets eliminate x, multiply theentire 1st equation by2.


45/64

45

Modeling Examples:

4 continued.


-2( x + y = 45 )

2x + 3y = 101

-2x + -2y = -90

2x + 3y = 101

Lets eliminate x, multiply theentire 1st equation by2.


46/64

46

Modeling Examples:

4 continued.


-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, sox = 34.

34 - 2 point shots and

11 - 3 point shots.


47/64

47

Modeling Examples:

5. Read problem 44 on page 363 in the textA Milk Mixture.


48/64

48

Modeling Examples:



let x = # gallons of 5% milk

let y = # gallons of skim (0%) milk


49/64

49

Modeling Examples:



let x = # gallons of 5% milk

let y = # gallons of skim (0%) milk


x + y = 100

This is because they want to make a mixturetotaling 100 gallons of milk.


50/64

50

Modeling Examples:

5. ContinuedWriting the 2nd equation:


51/64

51

Modeling Examples:


0.05x + 0.0y = 0.035(100)

Basically, we are multiplying the 1st equation bythe percent butterfat of the milk. Our finalmixture should be 3.5%, so we multiply0.035(100), since we want 100 total gallons.


52/64

52

Modeling Examples:

5. ContinuedThe two equations are:

x + y = 100

0.05x + 0.0y = 0.035(100)


53/64

53

Modeling Examples:


x + y = 100

0.05x + 0.0y = 0.035(100)

Next, multiply the 2nd equation by 1000 to removethe decimal. This gives us the following systemof equations: x + y = 100

50x + 0y = 35(100)


54/64

54

Modeling Examples:


x + y = 100

0.05x + 0.0y = 0.035(100)

Next, multiply the 2nd equation by 1000 to removethe decimal. This gives us the following systemof equations: x + y = 100

50x + 0y = 35(100)Solve the system (use substitution since the 2nd

equation has only one variable). The answerfollows on the next slide.


55/64

55

Modeling Examples:

5. ContinuedThe answer is 70 gallons of 5% milk and 30gallons of skim (0%) milk.


56/64

56

Modeling Examples:

6. Read problem 48 on page 363 in the textSchool Play Tickets.


57/64

57

Modeling Examples:



let x = # of adult tickets sold ($5 per ticket)

let y = # of student tickets sold ($2 per ticket)


58/64

58

Modeling Examples:



let x = # of adult tickets sold ($5 per ticket)

let y = # of student tickets sold ($2 per ticket)


x + y = 250

Since a total of 250 tickets were sold.


59/64

59

Modeling Examples:



60/64

60

Modeling Examples:


5x + 2y = 950

Basically, we multiplied the 1st equation by theprice of the tickets, and set it equal to theamount of money collected.


61/64

61

Modeling Examples:


5x + 2y = 950

Basically, we multiplied the 1st equation by theprice of the tickets, and set it equal to theamount of money collected.

Do you see how this is similar to example #4?

The 2 and 3 point shots?


62/64

62

Modeling Examples:

6. Continued


x + y = 250

5x + 2y = 950


63/64

63

Modeling Examples:

6. Continued


x + y = 250

5x + 2y = 950

Can you solve the system using eithersubstitution or addition? The answer follows onthe next slide.


64/64

Modeling Examples:

6. Continued

The answer is 150 adult tickets were sold, and100 student tickets were sold.

Documents

1_19_2016_Solving Systems of Linear Equations 2