Chapter 3 Linear Systems - TypePad · Systems of Linear Equations in Two Variables Solving Linear Systems by Graphing. There are three possible solutions to a system of linear equations

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Chapter 3 – Linear Systems

• Systems of Linear Equations

• Solving Systems of Equations by

Substitution

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:

1. Find ordered pairs that satisfy each of the

equations.

2. Plot the ordered pairs and sketch the

graphs of both equations on the same axis.

3. 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:

Consistent Dependent Inconsistent

One solution

Lines intersect

No solution

Lines are parallel

Infinite number of

solutions

Coincide-Same

line

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

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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!!

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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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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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.

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.

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 into

2nd equation and solve.

Step 3: Substitute –4 into 1st

equation and solve.

1

34

3

y

y

xy

The answer: ( -4 , -1)

1) Solve the system using substitution

x + y = 5

y = 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 = 5

2x = 2

x = 1


x + y = 5

y = 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 answer

would be if you graphed the two equations?


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

4x – 2y = 0

4(-3y + 7) – 2y = 0


3y + x = 7

4x – 2y = 0

Step 4: Plug back in to

find the other

variable.

4x – 2y = 0

4x – 2(2) = 0

4x – 4 = 0

4x = 4

x = 1

Step 3: Solve the

equation.

-12y + 28 – 2y = 0

-14y + 28 = 0

-14y = -28

y = 2


3y + x = 7

4x – 2y = 0

Step 5: Check your

solution.

(1, 2)

3(2) + (1) = 7

4(1) – 2(2) = 0

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

Substitution Method

Example Solve the system.

Solution

4

31

1

55

11623

11)3(23

3

y

y

x

x

xx

xx

xy 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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1123

yx

yx (1)

(2)

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)}


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.

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.

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.


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)}

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

3x – y = 4

x = 4y - 17

Your Turn:

Your Turn:

2x + 4y = 4

3x + 2y = 22

Clearing Fractions or Decimals

Systems without a Single Point Solution

0 = 4 untrue

Inconsistent Systems - how can you tell?

An inconsistent system has no solutions.

(parallel lines)

Substitution Technique

ntinconsiste

xxxx

xyBxyA

072225

332353

23)(53)(

0 = 0 or n = n

Dependent Systems – how can you tell?

A dependent system has infinitely many solutions.

(it’s the same line!) Substitution Technique

dependentyy

yyAyxyxB

xyBxyA

666633

6)3(23)(324128)(

24812)(623)(

23

23

34

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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:



First assign the variables:

let x = # of 2 point shots

let y = # of 3 point shots

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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

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Modeling Examples:

4 continued.

Writing the 2nd equation:

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Modeling Examples:

4 continued.


Some 2 pointers, some 3 pointers, for a total

score of 101 points

2x + 3y = 101

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Modeling Examples:

4 continued.


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.


-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.


-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.


-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.

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:


A Milk Mixture.


let x = # gallons of 5% milk

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

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Modeling Examples:


A Milk Mixture.


let x = # gallons of 5% milk

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


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


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Modeling Examples:

5. Continued


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


x + y = 100

0.05x + 0.0y = 0.035(100)

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Modeling Examples:

5. Continued


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


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:


School Play Tickets.

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Modeling Examples:




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

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

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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.

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Modeling Examples:

6. Continued


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Modeling Examples:

6. Continued


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


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


x + y = 250

5x + 2y = 950

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Modeling Examples:

6. Continued


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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Chapter 3 Linear Systems - TypePad · Systems of Linear Equations in Two Variables Solving Linear Systems by Graphing. There are three possible solutions to a system of linear equations