MATH 1324 Module 2 Notes: SYSTEMS OF EQUATIONS & MATRICES 2.0 Systems Of Equations … Mathematics/Chapter2... · MATH 1324 (Finite Mathematics or Business Math I) Lecture Notes Author:

MATH 1324 (Finite Mathematics or Business Math I) Lecture Notes Author: Kevin Pinegar

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MATH 1324 Module 2 Notes: SYSTEMS OF EQUATIONS & MATRICES

2.0 Systems Of Equations Review

A System of Linear Equations is a system that contains two or more linear equations. The systems will have multiple variables (usually x,y,z). A system could also use subscripted variables like x1,x2,x3. Obviously you could use other variables to fit what your variable represents (like t for time, or c for cost). Below are 3 examples of a system of linear equations

42

32

yx

yx

This is said to be a 2x2 system (2 equations, 2 variables)

31632

11

321

321

xxx

xxx


52

6352

143

zyx

zyx

zyx


Note: In practical applications, systems of equations could be very large. Consider all the variables require to launch a rocket (time, velocity, temperature, direction x, direction y, direction z, acceleration, momentum, friction, air resistance, etc..) In this lecture series I will mostly work with smaller systems like those above.

3 methods we will use to solve systems of equations without using matrices.

Graphing Method (graph the equations)

Substitution Method

Elimination (Addition) Method

Part I) Graphing Method For Solving Linear Systems With Only Two Variables (lines)

Graph each line by using intercepts or plotting points

Identify the intersection of the two lines (if it exists)

Where the lines intersect represent the solution (or set of solutions) to the system

1. Solve the system by graphing.

93

82

yx

yx

Equation y-int. x-int.

2x+y=8 (0,8) (4,0)

x+3y=9 (0,3) (9,0)

The solution is the point where the 2 graphs intersect. (3,2) Check the solution to make sure.


2

When solving a system involving only two variables, there are 3 distinct possibilities.

The lines could intersect at one point, giving a unique solution. o In this case, the system is said to be consistent and independent.

The lines could intersect at infinitely many points (same line is graphed) o In this case, the system is said to be consistent and dependent.

The lines may not intersect at all, giving no solution (parallel lines) o In this case, the system is said to be inconsistent.

You can also graph lines using the slope and y-intercept.

93

82

yx

yxSee above

2x+y=8 y=-2x+8

x+3y=16 y=(-1/3)x+(16/3) Different slope, Different y-intercept= Intersecting lines at single point

1624

82

yx

yx

2x+y=8 y=-2x+8

4x+2y=16 y=-2x+8 Same slope, same y-intercept= Same line

624

82

yx

yx

2x+y=8 y=-2x+8

4x+2y=6 y=-2x+3 Same slope, different y-intercept= Parallel lines

Unique Solution consistent and independent system

Infinite Solutions consistent and dependent system

No Solution inconsistent system

2:

1

3

yx

yxSolve by graphing

Equations x-intercept y-intercept

3 yx (3,0) (0,-3)

1 yx (-1,0) (0,-1)

The solution is the point (1,-2) Check!!!


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Vertical and Horizontal Lines

The graph of x=k where k is a constant is a vertical line

The graph of y=k where k is a constant is a horizontal line.

The equation x=-3 is a vertical line containing the point (-3,0) The equation y=-3 is a horizontal line containing the point (0,-3)

3:

1

3

y

yxSolve by graphing

Equations x-intercept y-intercept

3 yx (3,0) (0,-3)

1y none (0,-1)

The solution is the point (2,-1) Check!!!


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Part II) Substitution Method For Solving Linear Systems With Only Two Variables (lines)

Solve one of the equations for x or y in terms of the other variable

Substitute the expression into the other equation to solve for the unknown

Don’t forget to solve for the second variable.

1. Solve the system by substitution.

93

82

yx

yx

Solve the first equation for y:

82

82

xy

yx

Substitute into second equation:

3

155

9246

9823

x

x

xx

xx

Solve for the other variable:

28)3(2

82

y

xy

The solution (3,2) Check the solution to make sure.

2. Solve the system by substitution.

1

3

yx

yx

Solve the first equation for x:

3

3

yx

yx

Substitute into second equation:

2

42

132

13

y

y

y

yy

Solve for the other variable:

132

3

x

yx

The solution (1,-2) Check the solution to make sure.

In An Application Problem

A concert is selling adult tickets and child tickets to a concert. Suppose a person buys 2 adult tickets and 1 child ticket for $8. A second person paid $9 for 1 adult ticket and 3 child tickets. Find the price of the adult and child tickets.

Solution: Let x=price of an adult ticket and y=price of a child ticket.

First person bought 2 adult and one child First person paid $8 2x+1y=8 is total paid

Second person bought 1 adult and three child Second person paid $8 1x+3y=8 is total paid

See above problem.

93

82

yx

yx Solution: x=$3, y=$2

Kevin.Pinegar

Text Box

9


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Dependent And Inconsistent Systems

A Dependent System An Inconsistent System

3156

152

yx

yxSolve for x in terms of y.

2

1

2

5

152

152

yx

yx

yx

Substitute into other equation.

33

315315

3152

1

2

56

yy

yy

3=3 is an identity. A statement that is always true.So this system has infinite solutions.

3156

152

yx

yx Solve for x in terms of y.

2

1

2

5

152

152

yx

yx

yx

Substitute into other equation.

33

315315

3152

1

2

56

yy

yy

-3=3 is an impossible equation. A statement that is never true. So this system has no solution.

Application: Solving For the Break-Even Point (Revenue / Cost)

Revenue is the amount of money one takes in when selling products. If x represents the number of products sold and p represents the price, then the formula for revenue is R=xp. The price can be a constant like $45 per item, but sometimes the price is dependent on the number of items sold (p=45-0.01x).Since we are focusing on linear functions, our price will be constant.

Cost is the cost for a business to manufacturer goods based on the number x manufactured.. Cost usually has two parts.

o Fixed Cost is the cost regardless of the number of items manufactured (operating cost). Fixed cost is generally a constant like $30,000.

o Variable Cost is the cost based on the number of items (x) manufactured. For example, if it costs $25 to manufacture and item then the variable cost is 25x.

Cost Function is the fixed cost + variable cost. Example C= 30,000+25x

Break-Even Point is the point at which revenue and cost are equal.

Example: Suppose a company sells calculators for a price of $45. The weekly costs for the company is $30,000 plus $25 per calculator. Find the break-even point. Solution: The revenue function is R=45x. The cost function is C=30,000+25x. To find the break-even point, we need to find the value of x where R=C. To do this we will start with the cost function, C=30,000+25x and substitute the revenue 45x in place of R. Then solve for x. 45x=30000+25x 20x=30000 x=1500 So we must manufacture 1500 calculators per week to cover our costs. If we manufacture less than 1500, we experience a loss. If we manufacture more, we experience a profit. If you want to know the dollar value at the break-even point, plug x=2000 into either the revenue or cost function. R(1500)=45(1500)=$67,500. So break-even point is (1500,67500). See graphs:


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Graphs For C=30000+25x and R=45x

Application: Solving For the Equilibrium Point (Supply / Demand)

Demand: The quantity of a product that consumers are willing to purchase during a time period depends on its price. The higher the price, the lower the demand. Lowering the price increases the demand.

Supply: The quantity of a product that a supplier is willing to sell during a time period also depends on the price. The higher the price, more will be offered by the supplier. Lowering the price decreases the number the supplier is willing to supply.

These equations use x or q to represent quantity and use p to represent price. The quantity is the domain (x-axis) and the price is the range (y=axis). This can be confusing to explain because when we talk about increasing or decreasing price, we are referring to the y-axis.

Think of it another way: For demand, if the quantity increases then the price must be decreasing (negative slope). For supply, if the quantity increases then the price must be also increasing (positive slope).

o Sample Demand Equation: p= -0.5x+10 or p= -0.5q+10 o Sample Supply Equation: p= 0.03x+1.35 or p=0.03q+1.35

Equilibrium Point: The point where supply meets demand. (supply = demand)

Example: Suppose we want to analyze the sale of a box of tea where q represents the number of boxes of tea (in thousands) during one week. We have determined the following supply and

demand equations: Supply: 76.007.0 qp Demand: 42.0 qp

Find the equilibrium point.

Solution: Set supply (0.07q+0.76) equal to demand (-0.2q+4)

12

24.327.0

76.042.007.0

42.076.007.0

q

q

qq

qq

So if we provide 12,000 boxes of tea, consumers will consume that amount (theoretically). We need to set the price to:

60.1$76.0)12(07.076.007.0 qp

Equilibrium Point is (12,1.60). Let’s view the graph.


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Graph of Supply: p=0.07q+0.76 and Demand: p=-0.2q+4

Finding the supply and demand equations from data. Suppose when an item is priced $1.82 the supplier is willing to supply 9400 million bushels and the consumer consumes 9500. When the price increases to $1.94, the supplier will supply 9800 million bushels but the consumer consumes only 9300. What are the supply and demand equations?

Supply: You are given 2 points (x,p) Demand: You are given 2 points (x,p)

(9400,1.82) and (9800,1.94)

0003.094009800

82.194.1

m

bmxp (we can solve for b)

1

94.294.1

)9400(0003.094.1

b

b

b

Supply Equation: 10003.0 xp

(9500,1.82), (9300,1.94)

0006.095009300

82.194.1

m

bmxp (we can solve for b)

52.7

58.594.1

)9300(0006.094.1

b

b

b

Demand Equation: 52.70006.0 xp

Show the equilibrium point is (9466.67, 1.84)

Kevin.Pinegar

Pencil

Kevin.Pinegar

Text Box

9800


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Part III) Elimination Method (sometimes called the Addition Method)

If necessary multiply one or both equations by a constant that gives you coefficients of either x that are opposites or coefficients of either x that are opposites.

Add the corresponding terms of first equation to the second equation

One variable will eliminate and you can solve for the other.

Make sure you solve for the eliminated variable and write solution as an ordered pair.

1. Easy

1

3

yx

yx

Add the two equations and solve for x.

1

202

1

3)(

x

x

yx

yx

Use x to solve for y.

2

2

31

3

y

y

y

yx

Write answer as (1,-2)

2. Medium

93

82

yx

yx

Multiply the first equation by -3 (or the second equation by -2). I will do the latter to show you it does not matter which variable you solve for first.

)9(232

82

yx

yxor

1862

82

yx

yx

Add the two equations and solve for y.

2

105

y

y

Use y to solve for x.

3

62

822

82

x

x

x

yx

Write answer as (3,2)

3. Difficult

152

823

yx

yx

Multiply both equations by a constant. You have two options, but I will do the following. Multiply first equation by 2 and second equation by -3.

)1(3523

)8(2232

yx

yx

Add the results to solve for y.

1

1919

3156

1646

y

y

yx

yx

(continued in next cell)

3. Continued Use y to solve for x.

2

823

8)1(23

823

x

x

x

yx

Write the answer as (2,-1) Notes: Just like in the substitution method, if you get an identity equation there are infinite solutions ~ if you get an impossible equation there is no solution.


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Application: A man wants to use milk and juice to increase the amount of calcium and vitamin A in his daily diet. An oz. of milk contains 37 milligrams of calcium and 57 micrograms of vitamin A. An oz. of juice contains 5 milligrams of calcium and 65 micrograms of vitamin A. How many oz. of each should he consume daily to provide exactly 500 milligrams of calcium and 1200 micrograms of vitamin A?

Solution: x= number of oz. of milk. y=number of oz. of juice. See table

Milk Juice Needed

Calcium 37 5 500

Vitamin A 57 65 1200

Calcium Equation: 37x + 5y = 500 Vitamin A Equation: 57x + 65y = 1200

12006557

)500(1353713

yx

yx

12006557

650065481

yx

yx

Adding Gives:

.5.12

5300424

ozx

x

.5.7

5.375

50055.462

5005)5.12(37

ozy

y

y

y

12.5 oz. of milk and 7.5 oz. of juice.

They aren’t always nice friendly numbers. See example below.

243

1035

yx

yx

2433

10354

yx

yx

6129

401220)(

yx

yx

11/46

4611

x

x

Solve for x:

11/40

11

1203

11

230

11

1103

103)11/230(

103)11/46(5

y

y

y

y

y

Solution is (46/11,-40/11)


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Applications {These will be setup only, but you should be able to solve them if asked.}

1) A play has 500 seats. Child tickets cost $2 and adult tickets cost $3. If a sold out performance brings in revenue of $1350, how many child and adult tickets were sold? Solution: Let x represent the number of child tickets and y the number of adult tickets.

135032

500

yx

yx

2) A manufacturer produces two types of bicycles, 3-speed and 5-speed. It takes 1 hour to assemble a 3-speed bike, and 2 hours to assemble a 5-speed bike. It also takes 2 hours to paint a 3-speed bike and 3 hours to paint a 5-speed bike. There are only 20 hours available during a day to assemble bikes and 33 hours available to paint bikes. How many bike of each type must be produced each day in order to utilize all of the assembly and painting time? Solution: Let x = the number of 3-speed bikes manufactured and y= the number 5-speed bikes manufactured.

3332

202

yx

yx

A dependent system:

1536

52

yx

yx

1536

523

yx

yx

1536

1536)(

yx

yx 00

Dependent System: Both equations represent the same equation.

2

5

2

15252 yxyxyx

Solution set:

yy ,2

5

2

1 or

kk ,

2

5

2

1 where k is any real number.

{More on dependent systems later} A Note About Calculators: You can search online to find ways to use a calculator to solve and graph systems of equations.

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MATH 1324 Module 2 Notes: SYSTEMS OF EQUATIONS & MATRICES 2.0 Systems Of Equations … Mathematics/Chapter2... · MATH 1324 (Finite Mathematics or Business Math I) Lecture Notes Author: