INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith

Eric Smith

Applications of Systems of Equations

Slide Duration:

Section 1: Properties of Real Numbers
Basic Types of Numbers

30m 41s

Intro
0:00
Objectives
0:07
Basic Types of Numbers
0:36
Natural Numbers
1:02
Whole Numbers
1:29
Integers
2:04
Rational Numbers
2:38
Irrational Numbers
5:06
Imaginary Numbers
6:48
Basic Types of Numbers Cont.
8:09
The Big Picture
8:10
Real vs. Imaginary Numbers
8:30
Rational vs. Irrational Numbers
8:48
Basic Types of Numbers Cont.
10:55
Number Line
11:06
Absolute Value
11:44
Inequalities
12:39
Example 1
13:16
Example 2
17:30
Example 3
21:56
Example 4
24:27
Example 5
27:48
Operations on Numbers

19m 26s

Intro
0:00
Objectives
0:06
Operations on Numbers
0:25
0:53
Subtraction
1:33
Multiplication & Division
2:19
Exponents
3:24
Bases
4:04
Square Roots
4:59
Principle Square Roots
5:09
Perfect Squares
6:32
Simplifying and Combining Roots
6:52
Example 1
8:16
Example 2
12:30
Example 3
14:02
Example 4
16:27
Order of Operations

12m 6s

Intro
0:00
Objectives
0:06
The Order of Operations
0:25
Work Inside Parentheses
0:42
Simplify Exponents
0:52
Multiplication & Division from Left to Right
0:57
Addition & Subtraction from Left to Right
1:11
Remember PEMDAS
1:21
The Order of Operations Cont.
2:27
Example
2:43
Example 1
3:55
Example 2
5:36
Example 3
7:35
Example 4
8:56
Properties of Real Numbers

18m 52s

Intro
0:00
Objectives
0:07
The Properties of Real Numbers
0:23
Commutative Property of Addition and Multiplication
0:44
Associative Property of Addition and Multiplication
1:50
Distributive Property of Multiplication Over Addition
3:20
Division Property of Zero
4:46
Division Property of One
5:23
Multiplication Property of Zero
5:56
Multiplication Property of One
6:17
6:29
Why Are These Properties Important?
6:53
Example 1
9:16
Example 2
13:04
Example 3
14:30
Example 4
16:57
Section 2: Linear Equations
The Vocabulary of Linear Equations

12m 22s

Intro
0:00
Objectives
0:09
The Vocabulary of Linear Equations
0:44
Variables
0:52
Terms
1:09
Coefficients
1:40
Like Terms
2:18
Examples of Like Terms
2:37
Expressions
4:01
Equations
4:26
Linear Equations
5:04
Solutions
5:55
Example 1
6:16
Example 2
7:16
Example 3
8:45
Example 4
10:20
Solving Linear Equations in One Variable

28m 52s

Intro
0:00
Objectives
0:08
Solving Linear Equations in One Variable
0:34
Conditional Cases
0:51
Identity Cases
1:09
1:30
Solving Linear Equations in One Variable Cont.
2:00
2:10
Multiplication Property of Equality
2:43
Steps to Solve Linear Equations
3:14
Example 1
4:22
Example 2
8:21
Example 3
12:32
Example 4
14:19
Example 5
17:25
Example 6
22:17
Solving Formulas

12m 2s

Intro
0:00
Objectives
0:06
Solving Formulas
0:18
Formulas
0:26
Use the Same Properties as Solving Linear Equations
1:36
1:55
Multiplication Property of Equality
1:58
Steps to Solve Formulas
2:43
Example 1
3:56
Example 2
6:09
Example 3
8:39
Applications of Linear Equations

28m 41s

Intro
0:00
Objectives
0:10
Applications of Linear Equations
0:43
The Six-Step Method to Solving Word Problems
0:55
Common Terms
3:12
Example 1
5:03
Example 2
9:40
Example 3
13:48
Example 4
17:58
Example 5
23:28
Applications of Linear Equations, Motion & Mixtures

24m 26s

Intro
0:00
Objectives
0:21
Motion and Mixtures
0:46
Motion Problems: Distance, Rate, and Time
1:06
Mixture Problems: Amount, Percent, and Total
1:27
The Table Method
1:58
The Beaker Method
3:38
Example 1
5:05
Example 2
9:44
Example 3
14:20
Example 4
19:13
Section 3: Graphing
Rectangular Coordinate System

22m 55s

Intro
0:00
Objectives
0:11
The Rectangular Coordinate System
0:39
The Cartesian Coordinate System
0:40
X-Axis
0:54
Y-Axis
1:04
Origin
1:11
1:26
Ordered Pairs
2:10
Example 1
2:55
The Rectangular Coordinate System Cont.
6:09
X-Intercept
6:45
Y-Intercept
6:55
Relation of X-Values and Y-Values
7:30
Example 2
11:03
Example 3
12:13
Example 4
14:10
Example 5
18:38
Slope & Graphing

27m 58s

Intro
0:00
Objectives
0:11
Slope and Graphing
0:48
Standard Form
1:14
Example 1
2:24
Slope and Graphing Cont.
4:58
Slope, m
5:07
Slope is Rise over Run
6:11
Don't Mix Up the Coordinates
8:20
Example 2
9:39
Slope and Graphing Cont.
14:26
Slope-Intercept Form
14:34
Example 3
16:55
Example 4
18:00
Slope and Graphing Cont.
19:00
Rewriting an Equation in Slope-Intercept Form
19:39
Rewriting an Equation in Standard Form
20:09
Slopes of Vertical & Horizontal Lines
20:56
Example 5
22:49
Example 6
24:09
Example 7
25:59
Example 8
26:57
Linear Equations in Two Variables

20m 36s

Intro
0:00
Objectives
0:13
Linear Equations in Two Variables
0:36
Point-Slope Form
1:07
Substitute in the Point and the Slope
2:21
Parallel Lines: Two Lines with the Same Slope
4:05
Perpendicular Lines: Slopes are Negative Reciprocals of Each Other
4:39
Perpendicular Lines: Product of Slopes is -1
5:24
Example 1
6:02
Example 2
7:50
Example 3
10:49
Example 4
13:26
Example 5
15:30
Example 6
17:43
Section 4: Functions
Introduction to Functions

21m 24s

Intro
0:00
Objectives
0:07
Introduction to Functions
0:58
Relations
1:03
Functions
1:37
Independent Variables
2:00
Dependent Variables
2:11
Function Notation
2:21
Function
3:43
Input and Output
3:53
Introduction to Functions Cont.
4:45
Domain
4:46
Range
4:55
Functions Represented by a Diagram
6:41
Natural Domain
9:11
Evaluating Functions
12:02
Example 1
13:13
Example 2
15:03
Example 3
16:18
Example 4
19:54
Graphing Functions

16m 12s

Intro
0:00
Objectives
0:09
Graphing Functions
0:54
Using Slope-Intercept Form
1:56
Vertical Line Test
2:58
Determining the Domain
4:20
Determining the Range
5:43
Example 1
6:06
Example 2
7:18
Example 3
8:31
Example 4
11:04
Section 5: Systems of Linear Equations
Systems of Linear Equations

25m 54s

Intro
0:00
Objectives
0:13
Systems of Linear Equations
0:46
System of Equations
0:51
System of Linear Equations
1:15
Solutions
1:35
Points as Solutions
1:53
Finding Solutions Graphically
5:13
Example 1
6:37
Example 2
12:07
Systems of Linear Equations Cont.
17:01
One Solution, No Solution, or Infinite Solutions
17:10
Example 3
18:31
Example 4
22:37
Solving a System Using Substitution

20m 1s

Intro
0:00
Objectives
0:09
Solving a System Using Substitution
0:32
Substitution Method
1:24
Substitution Example
2:35
One Solution, No Solution, or Infinite Solutions
7:50
Example 1
9:45
Example 2
12:48
Example 3
15:01
Example 4
17:30
Solving a System Using Elimination

19m 40s

Intro
0:00
Objectives
0:09
Solving a System Using Elimination
0:27
Elimination Method
0:42
Elimination Example
2:01
One Solution, No Solution, or Infinite Solutions
7:05
Example 1
8:53
Example 2
11:46
Example 3
15:37
Example 4
17:45
Applications of Systems of Equations

24m 34s

Intro
0:00
Objectives
0:12
Applications of Systems of Equations
0:30
Word Problems
1:31
Example 1
2:17
Example 2
7:55
Example 3
13:07
Example 4
17:15
Section 6: Inequalities
Solving Linear Inequalities in One Variable

17m 13s

Intro
0:00
Objectives
0:08
Solving Linear Inequalities in One Variable
0:37
Inequality Expressions
0:46
Linear Inequality Solution Notations
3:40
Inequalities
3:51
Interval Notation
4:04
Number Lines
4:43
Set Builder Notation
5:24
Use Same Techniques as Solving Equations
6:59
'Flip' the Sign when Multiplying or Dividing by a Negative Number
7:12
'Flip' Example
7:50
Example 1
8:54
Example 2
11:40
Example 3
14:01
Compound Inequalities

16m 13s

Intro
0:00
Objectives
0:07
Compound Inequalities
0:37
'And' vs. 'Or'
0:44
'And'
3:24
'Or'
3:35
'And' Symbol, or Intersection
3:51
'Or' Symbol, or Union
4:13
Inequalities
4:41
Example 1
6:22
Example 2
9:30
Example 3
11:27
Example 4
13:49
Solving Equations with Absolute Values

14m 12s

Intro
0:00
Objectives
0:08
Solve Equations with Absolute Values
0:18
Solve Equations with Absolute Values Cont.
1:11
Steps to Solving Equations with Absolute Values
2:21
Example 1
3:23
Example 2
6:34
Example 3
10:12
Inequalities with Absolute Values

17m 7s

Intro
0:00
Objectives
0:07
Inequalities with Absolute Values
0:23
Recall…
2:08
Example 1
3:39
Example 2
6:06
Example 3
8:14
Example 4
10:29
Example 5
13:29
Graphing Inequalities in Two Variables

15m 33s

Intro
0:00
Objectives
0:07
Graphing Inequalities in Two Variables
0:32
Split Graph into Two Regions
1:53
Graphing Inequalities
5:44
Test Points
6:20
Example 1
7:11
Example 2
10:17
Example 3
13:06
Systems of Inequalities

21m 13s

Intro
0:00
Objectives
0:08
Systems of Inequalities
0:24
Test Points
1:10
Steps to Solve Systems of Inequalities
1:25
Example 1
2:23
Example 2
7:28
Example 3
12:51
Section 7: Polynomials
Integer Exponents

44m 51s

Intro
0:00
Objectives
0:09
Integer Exponents
0:42
Exponents 'Package' Multiplication
1:25
Example 1
2:00
Example 2
3:13
Integer Exponents Cont.
4:50
Product Rule for Exponents
4:51
Example 3
7:16
Example 4
10:15
Integer Exponents Cont.
13:13
Power Rule for Exponents
13:14
Power Rule with Multiplication and Division
15:33
Example 5
16:18
Integer Exponents Cont.
20:04
Example 6
20:41
Integer Exponents Cont.
25:52
Zero Exponent Rule
25:53
Quotient Rule
28:24
Negative Exponents
30:14
Negative Exponent Rule
32:27
Example 7
34:05
Example 8
36:15
Example 9
39:33
Example 10
43:16

18m 33s

Intro
0:00
Objectives
0:07
0:25
Terms
0:33
Coefficients
0:51
1:13
Like Terms
1:29
Polynomials
2:21
Monomials, Binomials, Trinomials, and Polynomials
5:41
Degrees
7:00
Evaluating Polynomials
8:12
9:25
Example 1
11:48
Example 2
13:00
Example 3
14:41
Example 4
16:15
Multiplying Polynomials

25m 7s

Intro
0:00
Objectives
0:06
Multiplying Polynomials
0:41
Distributive Property
1:00
Example 1
2:49
Multiplying Polynomials Cont.
8:22
Organize Terms with a Table
8:23
Example 2
13:40
Multiplying Polynomials Cont.
16:33
Multiplying Binomials with FOIL
16:48
Example 3
18:49
Example 4
20:04
Example 5
21:42
Dividing Polynomials

44m 56s

Intro
0:00
Objectives
0:07
Dividing Polynomials
0:29
Dividing Polynomials by Monomials
2:10
Dividing Polynomials by Polynomials
2:59
Dividing Numbers
4:09
Dividing Polynomials Example
8:39
Example 1
12:35
Example 2
14:40
Example 3
16:45
Example 4
21:13
Example 5
24:33
Example 6
29:02
Dividing Polynomials with Synthetic Division Method
33:36
Example 7
38:43
Example 8
42:24
Section 8: Factoring Polynomials
Greatest Common Factor & Factor by Grouping

28m 27s

Intro
0:00
Objectives
0:09
Greatest Common Factor
0:31
Factoring
0:40
Greatest Common Factor (GCF)
1:48
GCF for Polynomials
3:28
Factoring Polynomials
6:45
Prime
8:21
Example 1
9:14
Factor by Grouping
14:30
Steps to Factor by Grouping
17:03
Example 2
17:43
Example 3
19:20
Example 4
20:41
Example 5
22:29
Example 6
26:11
Factoring Trinomials

21m 44s

Intro
0:00
Objectives
0:06
Factoring Trinomials
0:25
Recall FOIL
0:26
Factor a Trinomial by Reversing FOIL
1:52
Tips when Using Reverse FOIL
5:31
Example 1
7:04
Example 2
9:09
Example 3
11:15
Example 4
13:41
Factoring Trinomials Cont.
15:50
Example 5
18:42
Factoring Trinomials Using the AC Method

30m 9s

Intro
0:00
Objectives
0:08
Factoring Trinomials Using the AC Method
0:27
Factoring when Leading Term has Coefficient Other Than 1
1:07
Reversing FOIL
1:18
Example 1
1:46
Example 2
4:28
Factoring Trinomials Using the AC Method Cont.
7:45
The AC Method
8:03
Steps to Using the AC Method
8:19
Tips on Using the AC Method
9:29
Example 3
10:45
Example 4
16:50
Example 5
21:08
Example 6
24:58
Special Factoring Techniques

30m 14s

Intro
0:00
Objectives
0:07
Special Factoring Techniques
0:26
Difference of Squares
1:46
Perfect Square Trinomials
2:38
No Sum of Squares
3:32
Special Factoring Techniques Cont.
4:03
Difference of Squares Example
4:04
Perfect Square Trinomials Example
5:29
Example 1
7:31
Example 2
9:59
Example 3
11:47
Example 4
15:09
Special Factoring Techniques Cont.
19:07
Sum of Cubes and Difference of Cubes
19:08
Example 5
23:13
Example 6
26:12

23m 38s

Intro
0:00
Objectives
0:08
0:19
0:20
Zero Factor Property
1:39
Zero Factor Property Example
2:34
Example 1
4:00
Solving Quadratic Equations by Factoring Cont.
5:54
Example 2
7:28
Example 3
11:09
Example 4
14:22
Solving Quadratic Equations by Factoring Cont.
18:17
Higher Degree Polynomial Equations
18:18
Example 5
20:22

29m 27s

Intro
0:00
Objectives
0:12
0:29
Linear Factors
0:38
1:22
Principle of Square Roots
3:36
Completing the Square
4:50
Steps for Using Completing the Square
5:15
Completing the Square Works on All Quadratic Equations
6:41
7:28
Discriminants
8:25
10:11
Example 1
11:54
Example 2
13:03
Example 3
16:30
Example 4
21:29
Example 5
25:07

16m 47s

Intro
0:00
Objectives
0:08
0:24
Using a Substitution
0:53
U-Substitution
1:26
Example 1
2:07
Example 2
5:36
Example 3
8:31
Example 4
11:14

29m 4s

Intro
0:00
Objectives
0:09
0:35
Squared Variable
0:40
Principle of Square Roots
0:51
Example 1
1:09
Example 2
2:04
3:34
Example 3
4:42
Example 4
13:33
Example 5
20:50

26m 53s

Intro
0:00
Objectives
0:06
0:39
Axis of Symmetry
1:46
Vertex
2:12
Transformations
2:57
3:23
Example 1
5:06
Example 2
6:02
Example 3
9:07
11:26
Completing the Square
12:02
Vertex Shortcut
12:16
Example 4
13:49
Example 5
17:25
Example 6
20:07
Example 7
23:43
Polynomial Inequalities

21m 42s

Intro
0:00
Objectives
0:07
Polynomial Inequalities
0:30
Solving Polynomial Inequalities
1:20
Example 1
2:45
Polynomial Inequalities Cont.
5:12
Larger Polynomials
5:13
Positive or Negative Intervals
7:16
Example 2
9:01
Example 3
13:53
Section 10: Rational Equations
Multiply & Divide Rational Expressions

26m 41s

Intro
0:00
Objectives
0:09
Multiply and Divide Rational Expressions
0:44
Rational Numbers
0:55
Dividing by Zero
1:45
Canceling Extra Factors
2:43
Negative Signs in Fractions
4:52
Multiplying Fractions
6:26
Dividing Fractions
7:17
Example 1
8:04
Example 2
14:01
Example 3
16:23
Example 4
18:56
Example 5
22:43

20m 24s

Intro
0:00
Objectives
0:07
0:41
Common Denominators
0:52
Common Denominator Examples
1:14
Steps to Adding and Subtracting Rational Expressions
2:39
Example 1
3:34
Example 2
5:27
Adding and Subtracting Rational Expressions Cont.
6:57
Least Common Denominators
6:58
Transitioning from Fractions to Rational Expressions
9:08
Identifying Least Common Denominators for Rational Expressions
9:56
10:41
Example 3
11:19
Example 4
12:36
Example 5
15:08
Example 6
16:46
Complex Fractions

18m 23s

Intro
0:00
Objectives
0:09
Complex Fractions
0:37
Dividing to Simplify Complex Fractions
1:10
Example 1
2:03
Example 2
3:58
Complex Fractions Cont.
9:15
Using the Least Common Denominator to Simplify Complex Fractions
9:16
10:07
Example 3
10:42
Example 4
14:28
Solving Rational Equations

16m 24s

Intro
0:00
Objectives
0:07
Solving Rational Equations
0:23
Isolate the Specified Variable
1:23
Example 1
1:58
Example 2
5:00
Example 3
8:23
Example 4
13:25
Rational Inequalities

18m 54s

Intro
0:00
Objectives
0:06
Rational Inequalities
0:18
Testing Intervals for Rational Inequalities
0:38
Steps to Solving Rational Inequalities
1:05
Tips to Solving Rational Inequalities
2:27
Example 1
3:33
Example 2
12:21
Applications of Rational Expressions

20m 20s

Intro
0:00
Objectives
0:07
Applications of Rational Expressions
0:27
Work Problems
1:05
Example 1
2:58
Example 2
6:45
Example 3
13:17
Example 4
16:37
Variation & Proportion

27m 4s

Intro
0:00
Objectives
0:10
Variation and Proportion
0:34
Variation
0:35
Inverse Variation
1:01
Direct Variation
1:10
Setting Up Proportions
1:31
Example 1
2:27
Example 2
5:36
Variation and Proportion Cont.
8:29
Inverse Variation
8:30
Example 3
9:20
Variation and Proportion Cont.
12:41
Constant of Proportionality
12:42
Example 4
13:59
Variation and Proportion Cont.
16:17
Varies Directly as the nth Power
16:30
Varies Inversely as the nth Power
16:53
Varies Jointly
17:09
Combining Variation Models
17:36
Example 5
19:09
Example 6
22:10
Rational Exponents

14m 32s

Intro
0:00
Objectives
0:07
Rational Exponents
0:32
Power on Top, Root on Bottom
1:05
Example 1
1:37
Rational Exponents Cont.
4:04
Using Rules from Exponents for Radicals as Exponents
4:05
Combining Terms Under a Single Root
4:50
Example 2
5:21
Example 3
7:39
Example 4
11:23
Example 5
13:14
Simplify Rational Exponents

15m 12s

Intro
0:00
Objectives
0:07
Simplify Rational Exponents
0:25
0:26
Product Rule to Simplify Square Roots
1:11
1:42
Applications of Product and Quotient Rules
2:17
Higher Roots
2:48
Example 1
3:39
Example 2
6:35
Example 3
8:41
Example 4
11:09

17m 22s

Intro
0:00
Objectives
0:07
0:33
Like Terms
1:29
Bases and Exponents May be Different
2:02
Bases and Powers Must be Same when Adding and Subtracting
2:42
3:55
Example 1
4:47
Example 2
6:00
7:10
Simplify the Bases to Look the Same
7:25
Example 3
8:23
Example 4
11:45
Example 5
15:10

19m 24s

Intro
0:00
Objectives
0:08
0:25
0:26
1:11
Don’t Distribute Powers
2:54
4:25
Rationalizing Denominators
6:40
Example 1
7:22
Example 2
8:32
9:23
Rationalizing Denominators with Higher Roots
9:25
Example 3
10:51
Example 4
11:53
13:13
Rationalizing Denominators with Conjugates
13:14
Example 5
15:52
Example 6
17:25

15m 5s

Intro
0:00
Objectives
0:07
0:17
0:18
Isolate the Roots and Raise to Power
0:34
Example 1
1:13
Example 2
3:09
7:04
7:05
Example 3
7:54
Example 4
13:07
Complex Numbers

29m 16s

Intro
0:00
Objectives
0:06
Complex Numbers
1:05
Imaginary Numbers
1:08
Complex Numbers
2:27
Real Parts
2:48
Imaginary Parts
2:51
Commutative, Associative, and Distributive Properties
3:35
4:04
Multiplying Complex Numbers
6:16
Dividing Complex Numbers
8:59
Complex Conjugate
9:07
Simplifying Powers of i
14:34
Shortcut for Simplifying Powers of i
18:33
Example 1
21:14
Example 2
22:15
Example 3
23:38
Example 4
26:33
Bookmark & Share Embed

## Copy & Paste this embed code into your website’s HTML

Please ensure that your website editor is in text mode when you paste the code.
(In Wordpress, the mode button is on the top right corner.)
×
• - Allow users to view the embedded video in full-size.
Since this lesson is not free, only the preview will appear on your website.

• ## Related Books

### Applications of Systems of Equations

• Some word problems lead to the development of a system of equations.
• Identify both unknowns used so you can quickly reference them later.
• Separate situations are a clue that more than one equation can be used to connect the information.
• Once a system is developed, you may use the substitution or elimination method to solve it.
• Make sure you check to see if the solution makes sense in the context of the problem.

### Applications of Systems of Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:12
• Applications of Systems of Equations 0:30
• Word Problems
• Example 1 2:17
• Example 2 7:55
• Example 3 13:07
• Example 4 17:15

### Transcription: Applications of Systems of Equations

Welcome back to www.educator.com.0000

In this lesson we are going to look at applications of systems of equations.0003

Think about word problems involving the systems of equations that we saw earlier.0007

Our main goal will be looking at these word problems and figuring out how we can interpret them and build the system from there.0015

Here is a part of this that we will be actually looking at our solutions and be able to figure out how it fits in the context of that word problem.0022

A good way to think of how to start picking these apart is to take every little bit of information in there.0033

Say 2 different situations and create an equation for each of those situations.0039

You will notice some of these problems look similar to ones we did in the earlier applications of linear equations.0045

In that lesson we tried to write everything in terms of one variable.0051

Now that we know more about systems, we can use 2 variables to help us out and create an equation for each of the situations present.0055

It will actually make things a little bit neater.0062

Once we have built our system of equations, we are getting down to being able to solve that system.0067

We will use our techniques for solving a system such as elimination method or the substitution method.0072

We can also use tables like we did earlier to help you organize this information0079

but I will mainly focus on just the methods of substitution and elimination to tackle the systems of equations.0083

One very important part to remember is that all of these are going to be word problems.0094

Even though we get a solution like x = 3 and y = 5, we have to take those and interpret them and say what is x and y.0099

Always interpret these in the context of the problem.0108

One thing that will make this very easy and not so bad, is to go ahead and write down what your variables are at the very beginning.0112

In that way, once you are all done with the problem you can reference that and say okay my x represents this and my y represents this.0122

That way you can easily say here is how it fits in the context of the problem.0130

Let us jump into our examples and see how we can actually start creating a system of equations.0140

In this first one, we have a restaurant that needs to have 2 seat tables and 3 seat tables.0146

According to local fire code, it looks like the restaurants maximum occupancy is 46 costumers.0151

If the owners have hardly enough servers to handle 17 tables of customers, how many of each kind of table should they purchase?0158

Notice that we are under a few different constraints.0165

One has to do with simply the number of people that you can fit in this restaurant.0168

We want to make sure that we have no more than 46 customers.0173

Another restriction that we are under is the number of tables that our servers can handle and provide good service and we can only do 17 tables.0177

From the owner's perspective, we do not want to go less than these numbers.0187

After all, if we have less than 46 customers, we are not making as much money.0190

If you have less than 17 tables then we have servers that are not doing a lot of work.0195

We are going to try and aim to get these exact when we set up our equations.0200

Let us hunt down some unknowns and first write those down.0205

I'm going to say let x that will be the number of 2 seat tables.0210

Let y be the number of 3 seat tables.0228

We can use these unknowns to start connecting our information gathered.0243

In the first situation, we are going to focus on the number of people that we want to fit in this restaurant.0248

We are looking to put 46 people in this restaurant and it will come from seating them either at these 2 seat table or at the 3 seat table.0254

There will be 2 people for every 2 seat table so we can represent that by saying 2 × X.0264

There will be 3 people at every 3 seat table or 3y.0272

That expression right there just represents a number of people and we want it equal to 46 customers.0277

That is dealing with our people, maybe I will even label that people.0285

The other restriction is what we can do with our servers in handling all of these tables.0291

We want the total number of tables and that is our 2 seat tables and our 3 seat tables to equals 17.0298

You can say we have encapsulated all the information to one equation or the other.0311

We have actually set up the system and it is just a matter of going through and solving it.0316

You could use the elimination method or you could use the substitution method, both of them should get you to the same answer.0321

I’m going to go through the elimination method.0327

I will do this by taking the second equation here and we will multiply that one by -2.0331

Let us see what the result will be.0340

The first equation will remain unchanged and everything in the second one will be multiplied by -2.0341

We have done that, we can go ahead and combine both of those equations together.0365

You will notice that the x’s are canceling out since one is 2 and one is -2, that is exactly what we want with the elimination method.0371

3y + - 2y will give us 1y, and now we have 46 - 834 and that will give us 12.0379

One of the great thing is about writing down our variables at the very beginning is not only do I know that y = 12,0397

but I can interpret this as the number of 3 seat tables since I have it right here that y is the number of 3 seat tables.0402

We are going to continue on, and I will figure out what x is so we can figure out the number for both of the tables.0410

It is not so bad, just always remember that the total number of tables must be equal to 17.0415

I will borrow one of our original equations down here and just substitute in the 12 for y.0421

To solve that one we will subtract 12 from both sides and get that x = 5.0433

Now we can fully answer this problem.0441

The number of 3 seat tables is 12 and the number of 2 seat tables that will be 5.0446

We will keep that as our final answer.0462

Set down your unknowns, set up your system, solve it and interpret it in the context of the problem.0468

Let us try another problem, it has a lot of the same feel to it.0478

This one says that at a concession stand, if you buy 5 hotdogs and 2 hamburgers the cost is $9.50, 2 hotdogs and 5 hamburgers cost$13.25.0483

We are interested in finding the total cost of just one hamburger.0495

First, we need to go ahead and label our unknown.0499

We have the number of hotdogs and we have our hamburgers.0504

Let us work on that.0509

I will say that we will use and see d for hotdogs and let us use h for our hamburgers.0512

Both of these situations here are involving costs.0529

In the first one, we are looking at so many hotdogs, 5 of them and 2 hamburgers all equal to $9.50.0534 In the next situation, we have a different combination of hotdogs and hamburgers to equal the$13.25.0543

We will take each of these and package them into their own equation.0549

5 hotdogs 2 hamburgers equals to $9.50.0554 2 hotdogs 5 hamburgers equals to$13.25.0569

There is our system of equations.0581

When we are done solving we should be able to get the price for just one hotdog and a price for just one hamburger.0583

In this system, you could solve it using elimination or substitution.0591

I think I'm going to attack this using our elimination method.0595

I will have to multiply both equations by something to get something to cancel out.0599

Let us multiply the first one here or multiply everything in there by -2.0604

We will take our second equation and I will multiply everything there by 5.0614

Let us see what the result is, -10 - 4h -19.0621

Then we will take our second equation 10 d + 25h (5 × 13.25).0639

Adding these 2 equations together, let us see what our result is.0661

The 10 d and -10 d both will cancel each other out and we will have 25h - 4h = 21h.0665

We can add together the 66 + 25 -19.0679

In this new equation, we only have h to worry about and we can get that all by itself just by dividing both sides by 21.0689

This will give me that h = $2.25.0702 By looking at what we have identified earlier, I know that this is the cost of one hamburger$2.25.0708

Even though the problem is not asking for it, let us go ahead and figure out how much a hotdog is0715

by taking this amount and substituting it back into one of our original equations.0719

You can see that I have put this into the first equation, now I need to multiply 2 by $2.25 that will$4.50.0737

Subtract $4.50 from both sides and divide both sides by 5.0750 Now we have our entire solution.0768 I know for sure that the hamburgers will cost$2.25.0769

I also know that one hotdog will cost exactly $1.0774 Try and pick out each situation there and interpret it into its own equation.0780 Let us try this one out.0789 In the U.S. Senate, there are 100 representatives, if there are exactly 10 more democrats than republicans, how many of each are in the senate?0791 One assumption that we are going to use here is that all representatives are either democrats or republicans.0799 We are not even going to consider any third parties here.0804 Let us see if I can break this down but first let us identify some variables.0808 Let us say r is the number of republicans and we will let d be the number of democrats.0813 We need to set up an equation for each of these and one of the first big bits of information I get is that the total number of representatives should be 100.0846 We can interpret that as the number of republicans + the number of democrats that should be equal to 100.0856 We need to interpret that there are exactly 10 more democrats than republicans.0870 To look at that, let us compare the 2.0879 If I was to take republicans and democrats and put them on each side of the equal sign,0881 they would not exactly be equal because I have 10 more democrats than republicans.0887 The way you want to interpret is that right now this side got a lot more, it has 10 more than the republican side.0893 If I want to balance out this equation, I need to take something away from the heavier side.0901 Since it is exactly 10 more, I will take away 10 and that should balance it just fine.0906 My second equation is r = d -10.0914 I will get that relationship that there are exactly 10 more democrats than republicans.0917 When we are done setting up the system, notice how it set up a good for substitution.0924 The reason is if you notice here our r is already solved.0929 Let us use that substitution method to help us out.0940 We will take the d – 10 and substitute it into the first equation.0943 In this new equation, we only have these to worry about.0958 Let us combine the like terms, 1d and another d will give us a total of 2d -10 =100 and we can add 10 to both sides.0961 One last step, let us go ahead and divide both sides by 2.0979 This will give me that the number of democrats should be 55.0987 Since I also know that there are exactly 10 more democrats than republicans,0992 I can take this number then subtract 10 and that will give me my republicans.0997 Let us say we have 55 democrats, 45 republicans.1011 I have one more example with these systems of equations and this involves a little bit more difficult numbers.1038 We are going to end up rounding these nice whole numbers, so we do not have to deal with too much decimals.1045 Be careful along the way because some of the numbers do get a little messy.1050 In this problem, I have a Janet that blends coffee for a coffee house1054 and what she wants to do is she wants to prepare 280 pounds of blended coffee beans and sell it for$5.32 per pound.1058

The way she plans on making this mixture is she is going to blend together 2 different types of coffees.1066

She is going to blend together a high-quality coffee that cost $6.25/pound.1072 And she is also going to blend together with that a cheaper coffee that only costs$3/ pound.1080

The question is to the nearest pound, how much high quality coffee bean and how much cheaper quality coffee bean should she mix in order to get this plan.1088

Let us set down some unknowns and see if we can figure this out.1098

I will call (q) the number of pounds and this will be for our high quality coffee.1105

Let (c), even number of pounds for our cheap quality coffee.1126

We are going to try and set up a how many pounds of each of these we need.1149

One of the first bits of information that will help out with that is we know that we have a total of 280 pounds.1153

The number of pounds for my high quality coffee + the number of pounds for my cheaper coffee better equal to 280 pounds.1160

This entire equation just deals with pounds.1170

The second part will have to deal with the cost.1176

We want the final mixture to be $5.32, so we will use the cost of the high quality coffee bean and the cost of the low quality coffee bean1178 and blend the 2 together and get that final cost.1186 Let us see what we have here.1191 Normally, high quality coffee costs$6.25 every pound of high quality coffee.1191

The cheaper coffee usually costs $3/pound.1200 We are hoping to make is$5.32 for that final 280 pounds of coffee.1206

Let us mark this other one as just costs.1215

In our system here, you can see the numbers are not quite as nice but we can definitely still work with them.1221

Let us do a little bit of cleaning up then we will go ahead and try the substitution method on the system.1227

I’m going to multiply these together first, so $5.32 × 280,1234 I'm doing the new system 625q +$3, c =1489.6.1248

To use the substitution method, I’m going to take the first equation and let us go ahead and solve it for q.1265

Q = 280 – c.1274

Once we have that, we can go ahead and substitute it into our second equation.1280

$6.25 q is 280 – (c +$3.04c) = 1489.6.1289

Our new equation only has c and we can go ahead and solve for that.1307

Let us distribute through by the \$6.25 and see what we will get.1312

17.50 – 6.25c + 3c = 1489.6.1323

Let us combine our c terms, -3.25c = 1489.6.1338

We almost have c all by itself, let us get there by subtracting the 17.50 from both sides, -260.4.1354

One last step, let us go ahead and divide both sides by the -3.25.1378

I’m getting 80 and then something after the decimal 12307692, it just keeps going on and on.1396

I can see that I’m going to have just a little bit more than 80 pounds, but we want to round this to the nearest whole pound.1407

What I will say is that c is about equal to 80 pounds.1415

Now, once I know how many pound the cheaper coffee I will need, we can quickly go back to one of our originals1421

and figure out how much of the high quality coffee we need.1427

I need the total poundage to be 280 and 80 of that will be in the cheaper coffee, then I know that the high quality coffee I will need 200 pound.1433

Our rounded solution, I have that 80 pounds of cheap coffee and 200 pounds of high quality coffee.1446

If you pick apart your word problems and try and set down an equation for some of the situations in there,1454

you should be able to develop your system of equations just fine.1459

Then use one of the methods that we have learned especially those algebraic methods to solve the system1463

and do not forget to interpret your solution in the context of the problem.1467

Thanks for watching www.educator.com.1472

OR

### Start Learning Now

Our free lessons will get you started (Adobe Flash® required).