Copyright © 2008 Pearson Education Canada4-1 Chapter 4 Linear Systems Contemporary Business...

24
Copyright © 2008 Pearson Education Canada 4-1 Chapter 4 Linear Systems Contemporary Business Mathematics with Canadian Applications Eighth Edition S. A. Hummelbrunner/K. Suzanne Coombs PowerPoint: D. Johnston

Transcript of Copyright © 2008 Pearson Education Canada4-1 Chapter 4 Linear Systems Contemporary Business...

Copyright © 2008 Pearson Education Canada 4-1

Chapter 4

Linear Systems

Contemporary Business Mathematics with Canadian Applications

Eighth Edition S. A. Hummelbrunner/K. Suzanne Coombs

PowerPoint: D. Johnston

Copyright © 2008 Pearson Education Canada 4-2

ObjectivesAfter completing chapter four the student will

be able to:• Use of Linear System in deciding the combination of production

levels to use of labour, materials, and produce highest level of profit.

• Use of Linear System to decide how to best allocate business resources.

• Use the method of elimination to solve a linear system of two simultaneous equations in two variables.

• Graph linear equations.• Graph linear systems of two equations with two variables.• Solve problems using linear systems .

Copyright © 2008 Pearson Education Canada 4-3

Algebraic Elimination Method

Consider the following system of two linear equations with unknown variables X and Y. X + Y = 5 X - Y = 11 Note that the coefficients of Y are +1 and –1. Eliminate Y by adding : (X + Y) + (X – Y) = 5 + 11 2X = 16 or X = 8 (continued)

Copyright © 2008 Pearson Education Canada 4-4

Algebraic Elimination Method(continued)

Substitute X = 8 into the first equationX + Y =5.

This results in 8 + Y = 5 or Y = -3.

The solution is X = 8, y = -3. Substitute theresult back into the original system as a check.

Copyright © 2008 Pearson Education Canada 4-5

Unequal CoefficientsConsider the following system. X – 3Y = 10 3X + 2Y = 52

Multiply the first equation by -3 (both sides).

-3X + 9Y = -30 3X + 2Y = 52Add “equals to equals” to get 11 Y = 22 orY = 2. Substitute Y = 2 into first equation toget X = 16. The solution is X=16 and Y=2.

SOLVE

Copyright © 2008 Pearson Education Canada 4-6

1. y = 3x + 12 x = –y

2. 6x + 3y = 24 2x + 9y = –8

Copyright © 2008 Pearson Education Canada 4-7

Graphing Linear Equationsin a system of rectangular coordinates

Copyright © 2008 Pearson Education Canada 4-8

Locating a Point

Constructing a table of values

• To graph a linear equation a suitable set of points may be obtained by constructing a table of values.

• Substitute random value for X or Y and computer the value of the second variable.

• List ordered pairs (X,Y) in a table of values.

Copyright © 2008 Pearson Education Canada 4-9

Copyright © 2008 Pearson Education Canada 4-10

Constructing a Table for x + y = 4

or y = 4 - x

X 0 2 4 Y 4 2 0

SOLVE

Copyright © 2008 Pearson Education Canada 4-11

Construct a table of values forX = Y – 2 , for values of Y from -3 to +5

Copyright © 2008 Pearson Education Canada 4-12

Graphing an Equationx + y = 4

Copyright © 2008 Pearson Education Canada 4-13

Special CasesLines Parallel to X-axis

Note the graphs for y = 3 and y = -3.

Copyright © 2008 Pearson Education Canada 4-14

Special CasesLines Parallel to Y-axis

Note the graphs of x =3 and x = -3.

Copyright © 2008 Pearson Education Canada 4-15

Slope-y-intercept Form of a Linear Equation

Y = mx + b Linear equation

m Slope = rise run

b y-intercept

Y = 3X + 8

Slope = 3 y-intercept = 8

Copyright © 2008 Pearson Education Canada 4-16

Writing an Equation in Slope-y-intercept Form

x – 3y + 300 = 0 Original equation

-3y = -x - 300 Rearrange terms .

y = x + 100 3

Divide both sides by-3.

Slope is 1/3. Y-intercept is 100.

Copyright © 2008 Pearson Education Canada 4-17

Special Cases for the Slope

Lines parallel to x-axis O slope

Lines parallel to y-axis Undefined slope

Copyright © 2008 Pearson Education Canada 4-18

Different Cases for the Slope

Positive slope

Negative slope

O slope

Undefined slope

SOLVE

Copyright © 2008 Pearson Education Canada 4-19

Page 162

Exercise 4.2

Write the coordinatesA – 1

Plot the point in system of rectangular axesA – 2

Page 163

Exercise 4.2

Graph the equationsB – 1B – 2

Copyright © 2008 Pearson Education Canada 4-20

Graph a System of Linear Equations

x = y

x-2y +2000 = 0

Copyright © 2008 Pearson Education Canada 4-21

Problem SolvingHeather invested $15,000 in two mutual funds A and B. Her investment in mutual fund A is $5,000 greater than twice her investment in B. Find the amount invested in each fund.

X = amount in A Y = amount in B

X + Y = 15000 X – 2Y = 5000 (continued)

Copyright © 2008 Pearson Education Canada 4-22

Problem Solving (continued)

Consider the system generated by Heather’sinvestments. X + Y = 15000 X – 2Y = 5000 X can be eliminated by subtracting “equalsfrom equals”. This gives 3Y = 10,000 orY = 3333.33. Substitute this value for Y intothe first equation to get X = 11,666.67.

SOLVEfollowing Linear Systems

graphically

Copyright © 2008 Pearson Education Canada 4-23

1. X + Y = 4 and X – Y = - 4

2. 5X – 2Y = 20 and Y = 5

Copyright © 2008 Pearson Education Canada 4-24

Summary• Linear systems of two simultaneous

equations in two variables are useful for modelling business applications.

• Such systems can be solved by graphing the two linear equations and finding the point of intersection.

• Such systems can be solved algebraically by using the method of elimination.