W&C - Chapter 2

Copyright (c) 2004 Brooks/Cole, a division of Thomson Learning, Inc.

W&C - Chapter 2W&C - Chapter 2

Systems of Linear EquationsSystems of Linear Equations(essentially a review of what we did with breakeven (essentially a review of what we did with breakeven analysis, but better and more detailed in every way)analysis, but better and more detailed in every way)


Deja vuDeja vu

R(x) = 1.95xC(x) = .03x + 9000

We went through the graphical and table solutions.


solving algebraiclysolving algebraicly

The major change will be to write the functions this cool way:The major change will be to write the functions this cool way:ax + by = cax + by = c

The number The number aa is called the is called the coefficient of coefficient of xx.. The number The number bb is called the is called the coefficient of coefficient of yy..

xx and and yy are the are the variablesvariables that represent the unknowns. that represent the unknowns.

Example: Example: Revenue on drinks Revenue on drinks

y = 1.95 x + 0y = 1.95 x + 0 becomes becomes -1.95x + 1y = 0-1.95x + 1y = 0 Cost of drinksCost of drinks

y = .03 x + 9000y = .03 x + 9000 becomesbecomes -.03x + 1y = 9000-.03x + 1y = 9000

where where – 1.95 and -.03 – 1.95 and -.03 are considered are considered coefficients ofcoefficients of xx and and

1 and 11 and 1 are considered are considered coefficients ofcoefficients of yy..


A System of Two Linear Equations and Two UnknownsA System of Two Linear Equations and Two Unknowns

Because we worked with two functions at a time (Revenue and Because we worked with two functions at a time (Revenue and Cost functions), we get to claim we were working with a Cost functions), we get to claim we were working with a ““system of two linear equationssystem of two linear equations””

ax + by = cax + by = c

px + qy = r px + qy = r

where where x x and and yy are variables that represent the are variables that represent the two unknownstwo unknowns.. The values for The values for xx and and yy that satisfy that satisfy bothboth equations will be called equations will be called

the “the “solution solution for the system.” for the system.” When calculating a breakeven point (or when working with When calculating a breakeven point (or when working with

any system of linear equations), we get to call the breakeven any system of linear equations), we get to call the breakeven point the “solution for the system.”point the “solution for the system.”


Solution to the system of two linear equationsSolution to the system of two linear equations Revenue on drinks Revenue on drinks

-1.95x + 1y = 0-1.95x + 1y = 0 Cost of drinksCost of drinks

-.03x + 1y = 9000-.03x + 1y = 9000 Multiply by -1 and add the equations to eliminate yMultiply by -1 and add the equations to eliminate y

-1.95x + 1y = 0 -1.95x + 1y = 0 .03x - 1y = -9000 .03x - 1y = -9000 multiplied equation by -1 multiplied equation by -1

-1.92x = -9000-1.92x = -9000 x = 4687.5, the breakeven point x = 4687.5, the breakeven point

that’s half the “solution to the system”that’s half the “solution to the system” The other half is to calculate y The other half is to calculate y

y = 1.95xy = 1.95x y = 1.95(4687.5)y = 1.95(4687.5)

y = 9140.625y = 9140.625 Breakeven is therefore 4687.5 drinks, when revenue and cost are both Breakeven is therefore 4687.5 drinks, when revenue and cost are both

$9,140.625$9,140.625. . . Right?. . . Right?


Psychology problemPsychology problem

Suppose you have Suppose you have $3$3 to spend on snacks and a drink. to spend on snacks and a drink. If If xx represents the amount you’ll spend on represents the amount you’ll spend on snackssnacks and and yy the the

amount you’ll spend on a amount you’ll spend on a drinkdrink, you can say that, you can say thatx + y = x + y = 33

If you have obsessive-compulsive disorder and must spend exactly If you have obsessive-compulsive disorder and must spend exactly $1$1 more on snacks than on your drink, you can also say that: more on snacks than on your drink, you can also say that:

x - y = x - y = 11

These two equations represent a These two equations represent a system of equationssystem of equations.. The values for The values for xx and and yy that satisfy that satisfy bothboth equations are called the equations are called the

solution solution for the system.for the system. The solution can be obtained The solution can be obtained graphicallygraphically or or algebraicallyalgebraically..


Graphical SolutionGraphical Solution

One way to solve the system of equations is to One way to solve the system of equations is to graph the linesgraph the lines representing each equation and representing each equation and find the coordinatesfind the coordinates of the point at of the point at which the lines which the lines crosscross: :

x

y

x + y = x + y = 33

x - y = x - y = 11

(2, 1)

Thus, the optimal Thus, the optimal solution is:solution is:

x x = 2= 2 ($2 spent on snacks)($2 spent on snacks)

y y = 1 = 1 ($1 spent on a drink)($1 spent on a drink)


In the algebraic approach, we try to In the algebraic approach, we try to combine the equationscombine the equations in in such a way as to such a way as to eliminate one variableeliminate one variable. .

One wayOne way of doing this is by of doing this is by adding the equationsadding the equations or or multiples multiples of the equationsof the equations to find the value of one variable. to find the value of one variable.

Then Then substitutingsubstituting in one of the equations to find the value of in one of the equations to find the value of the other variable.the other variable.

Algebraic Solution: Adding EquationsAlgebraic Solution: Adding Equations

xx ++ yy == 33x - y = 1

(2)(2) ++ yy == 33

x = 22x + 0 = 4

y = 3 - 2y = 1


Solving a System: Algebraically versus GraphicallySolving a System: Algebraically versus Graphically

The algebraic method always allows us to determine an The algebraic method always allows us to determine an exact exact solutionsolution, whereas the graphic method often does not., whereas the graphic method often does not.


Graphical solution problemsGraphical solution problems Solve the system:Solve the system:

33x + x + 55yy = = 0022x + x + 77yy = = 11

Graphic Method:Graphic Method:

x

y

22x + x + 7y7y = = 11

33x + x + 55y = y = 00

(-.5, -.3)?


Try it in pairs #1!Try it in pairs #1! Solve the system:Solve the system:

x – x – 33yy = = 55– – 22x + x + 66yy = = 88

Algebraic Method:Algebraic Method:



x – x – 33yy = = 55– – 22x + x + 66yy = = 88

Multiply equation one by Multiply equation one by 22 and add: and add:

22x x – 66yy = = 1010

– – 22x + x + 66yy = = 88 00 + + 00 = = 1818

00 = = 1818


x 2



x – x – 33yy = = 55– – 22x + x + 66yy = = 88

Multiply equation one by Multiply equation one by 22 and add: and add:

22x x – 66yy = = 1010

– – 22x + x + 66yy = = 88 00 + + 00 = = 1818

00 = = 1818

The system has The system has no solutionno solution and is called and is called inconsistentinconsistent..(The two lines are (The two lines are parallelparallel, so they , so they never meetnever meet))


x 2



x + yx + y = = 2222x + x + 22yy = = 44




x + yx + y = = 2222x + x + 22yy = = 44

Multiply equation one by Multiply equation one by (– 2)(– 2) and add: and add:

– – 22x x – 22yy = = – 4– 4

22x + 2yx + 2y = = 44 00 + + 00 = = 00

00 = = 00


x (– 2)



x + yx + y = = 2222x + x + 22yy = = 44

Multiply equation one by Multiply equation one by (– 2)(– 2) and add: and add:

– – 22x x – 22yy = = – 4– 4

22x + 2yx + 2y = = 44 00 + + 00 = = 00

00 = = 00

This is called a This is called a redundant systemredundant system:: the two equations are in fact the two equations are in fact two versions of the same equationtwo versions of the same equation..

When we have a single equation with two unknowns, there When we have a single equation with two unknowns, there are are infinitely many solutionsinfinitely many solutions (all the points along the line). (all the points along the line).


x (– 2)


Example: Example: Equilibrium PriceEquilibrium Price

The The demanddemand for refrigerators in West Podunk is given by for refrigerators in West Podunk is given by

where where qq is the number of refrigerators that the citizens will is the number of refrigerators that the citizens will buy each year if they are priced at buy each year if they are priced at pp dollars each. dollars each.

10010

pq



The The demanddemand for refrigerators in West Podunk is given by for refrigerators in West Podunk is given by

where where qq is the number of refrigerators that the citizens will is the number of refrigerators that the citizens will buy each year if they are priced at buy each year if they are priced at pp dollars each. dollars each.

The The supplysupply is is

where now where now qq is the number of refrigerators the manufacturers is the number of refrigerators the manufacturers will be willing to ship into town each year if they are priced at will be willing to ship into town each year if they are priced at pp dollars each. dollars each.

10010

pq

2520

pq

Find the Find the equilibrium priceequilibrium price and and quantityquantity..


125

100

75

50

25

0 100 300 500 700

Solve the system:Solve the system:

Graphic Method:Graphic Method:

p

q (500, 50)

10010

pq 25

20

pq

10010

pq

2520

pq

Solution:Solution: The equilibrium The equilibrium

price is $500.price is $500. The equilibrium The equilibrium

quantity is 50 quantity is 50 refrigerators refrigerators per year.per year.


Now solve the system Now solve the system using the algebraic using the algebraic methodmethod




Write both equations in Write both equations in standard formstandard form::

10010

pq

2520

pq

10010

pq

2520

pq




Write both equations in Write both equations in standard formstandard form::

Multiply the first equation by Multiply the first equation by 1010 and the second by and the second by 2020 to to clear clear fractionsfractions, then , then add and solveadd and solve::

11pp + 10 + 10qq = = 10001000– – 11pp + 20 + 20qq = = 500500

0p + 300p + 30qq = = 15001500qq = = 5050

10010

pq

2520

pq

10010

pq

2520

pq

Then Then substitutesubstitute q q = 50= 50 in in equation one to find equation one to find pp::

pp + 10(50) = + 10(50) = 10001000pp + 500 = + 500 = 10001000

pp = = 1000 – 1000 – 500500

pp = = 500500

x 10

x 20


11pp + 10 + 10qq = =10001000– – 11pp + 20 + 20qq = =500500

00pp + 30 + 30qq = =15001500 qq = = 5050

Something interesting here. We almost don’t even need p and q Something interesting here. We almost don’t even need p and q or the + or the = to do the math.or the + or the = to do the math.

1 10 10001 10 1000– – 1 20 5001 20 500

0 30 15000 30 1500I’m lazy.

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W&C - Chapter 2

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