Copyright © 2008 Pearson Education Canada4-1 Chapter 4 Linear Systems Contemporary Business...
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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
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 4-13
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.