Chabot Mathematics
§3.2b System
Applications
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Review §
3.2
MTH 55
 Any QUESTIONS About
• §3.2a → System Applications
 Any QUESTIONS About HomeWork
• §3.2a → HW-09
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Summary of Eqn Elimination
1. Arrange the equations with
like terms in columns.
2. Multiply one or both equations by an
appropriate factor so that the new coefficients
of x or y have the same absolute value.
3. Add or subtract the equations and solve for
the remaining variable.
4. Substitute the value for that variable into one
of the equations and solve for the value of the
other variable.
5. Check the solution in each of the
original equations.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Solve by Elimination
 Consider This
Equation System
x  4 y  17
3x  2 y  9
 If the top equation
was multiplied by 3,
then the first term
would be 3x. The
bottom equation
could then be
subtracted →
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 → from the top
equation eliminating
the variable x
3( x  4 y )  3(17)
3x  12 y  51
 Then the New
System of Eqns
3x  12 y  51
3x  2 y  9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Eqn Elimination cont.1
 Choose to Find y first; using Algebra:
3x  12 y  51
3x  2 y  9
14 y  42
14 y 42

14 14
y3
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Subtract the bottom equation
from the top equation.
Divide Both Sides by 14
Solve for y
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Eqn Elimination cont.2
 Solve for x by substituting the value for
y (y = 3) into one of the equations.
3x  2 y  9
3x  6  6  9  6
Add 6 to Both Sides
3 x  15
Combine Like Terms
3 x  15 1 3
x5
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3x  2( 3)  9
Mult Both Sides by 1/3
Solve for x
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Eqn Elimination cont.3
 Thus the Solution to the Eqn System
x  4 y  17
3x  2 y  9
x5
y3
 To Check, Substitute the value of the
variables into each original equation
x  4 y  17
3x  2 y  9
3( 5 )  2( 3)  9
5  4 ( 3)  17
5  12  17
15  6  9
17  17
99
Chabot College Mathematics
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

Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Graphically
7
6
x  4 y  17
5
(5,3)
4
3
2
1
0
-1
0
1
2
-1
3
4
5
6
7
8
9
3x  2 y  9
-2
-3
file = M65_§7-1_Graphs_0607.xls
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
The Problem Solving Cycle
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Applications Tips
 The Most Important Part of Solving
REAL WORLD (Applied Math) Problems
 The Two Keys to the Translation
• Use the LET Statement to ASSIGN
VARIABLES (Letters) to Unknown Quantities
• Analyze the RELATIONSHIP Among the
Variables and Constraints (Constants)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Motion Problems
 Recall the Motion Formula:
Distance = {Rate (or speed)}  {Time}
• Symbolically → d = r•t
 Game plan for Solving Motion Probs
• Draw a diagram using an arrow or arrows to
represent distance and the direction of each
object in motion.
• Organize the information in a table or chart.
• Look for as many things as you can that are
the same, so you can write equations
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
 Miguel’s motorboat took 4 hr to make a
trip downstream with a 5-mph current.
The return trip against the same current
took 6 hr. Find the speed of the boat in
still water.
A
QUICK
Diagram
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
1. Familiarize. Note that the current
speeds up the boat when going
downstream, but slows down the
boat when going upstream. For our
guess, suppose that the speed of the
boat with no current is 20 mph.
The boat would then travel
•
20 + 5 = 25 mph downstream
•
20 – 5 = 15 mph upstream.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
 Assess 20 mph Guess
• In 4 hr downstream the boat would travel
4(25) = 100 mi.
• In 6 hr upstream the boat would travel
6(15) = 90 mi.
 So our guess of
20 mph is incorrect,
but it seems
pretty close.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
2. Translate:



LET r ≡ the rate of the boat in still water.
Then

r + 5 = the boat’s speed downstream

r – 5 = the boat’s speed upstream.
Tabulate d = r•t calculations
Downstream
Upstream
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Distance
Rate
Time
d
d
r+5
r–5
4
6
d = (r + 5)4
d = (r – 5)6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
 The Table produced d  (r  5)4,
an Eqn System
d  (r  5)6.
3. Solve: Use Substitution
(r + 5)4 = (r – 5)6
4r + 20 = 6r – 30
50 = 2r
25 = r
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Boat in Motion
4. Check: When r = 25 mph, the speed
downstream is 30 mph and the speed
upstream is 20 mph. The distance
downstream is 30(4) = 120 mi and the
distance upstream is 20(6) = 120 mi,
so we have a check.
5. State: The speed of
the boat in still water
is 25 mph.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
 Tamika and Ernesto are traveling north
in separate cars on the same highway.
Tamika is traveling at 55 miles per hour
and Ernesto is traveling at 70 miles per
hour. Tamika passes Exit 54 at 1:30
p.m. Ernesto passes the same exit at
1:45 p.m.
 At what time will Ernesto
catch up with Tamika?
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
1. Familiarize: To determine what time
Ernesto will catch up with Tamika, we
need to calculate the amount of time it
will take him to catch up to her. We
can then add the amount to 1:45pm
2. Translate: LET
•
x ≡ Tamika’s travel time after
passing Exit 54
•
y ≡ Ernesto’s travel time after
passing Exit 54
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
 Tabulate (distance) = (spd)(time) calcs
Category
Tamika
Ernesto
Speed
55
70
Time
x
y
Distance
55x
70y
 Translate – Connect the Time Difference:
Ernesto reaches exit-54 15 minutes after
Tamika; Tamika will have
traveled 15 minutes
1
x

y

(¼ hr) longer →
4
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
 Translate: When Ernesto catches up,
they will have traveled the same
distance; i.e.; dTamika = dErnesto.
• From
Table
55x  70 y
3. Carry out: The
translations
produced a
System of Eqns
Chabot College Mathematics
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1
x y
4
55 x  70 y
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
 Solve by Substitution
 Sub y = 0.92 into first
Eqn to solve for x
1
x y
4
x  0.92  0.25
x  1.17
Chabot College Mathematics
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1
x y
4
55 x  70 y
1

55  y    70 y
4

55
55 y 
 70 y
4
13.75  15 y
0.92  y
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Autos in Motion
4. Check: Both Eqns
1
x y
4
1.17  0.92  0.25
1.17  1.17

55 x  70 y
55(1.17)  70(0.92)
64.35  64.4 
5. State: Ernesto will catch up to Tamika
in a little less than 1 hour (0.92 hrs,
which is 55 min). The time will be
1:45pm + (55 min) = 2:40pm
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Supply and Demand
 As the price of a product varies, the
amount sold varies. Consumers
will demand less as price goes up.
Sellers will
supply more
as the price
goes up.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Supply & Demand Equilibrium
Supply
Quantity
Equilibrium point
Demand
Price
 The point of intersection is called the
equilibrium point. At that price, the amount
that the sellers will supply is the same amount
that the consumers will buy
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Supply = Demand
 Find the equilibrium point if the supply
and demand functions for a new brand
of digital video recorder (DVR) are given
by the system
supply
demand
p  60  0.0012x
p  80  0.0008x
 Where
• p is the unit-price in dollars
• x is the number of units
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
(1)
(2)
Example  Supply = Demand
1. Familiarize: The word “Equilibrium” in
Supply & Demand problems means that
Supply & Demand are Exactly the
Same.
2. Translate: S = D →
p D  80  0.0008x  60  0.012 x  pS
3. Carry Out: Isolate x in the above Eqn to
find the quantity. Then Substitute the
value of x into either Price Eqn
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Supply = Demand
 Subbing p  80  0.0008x
60  0.0012x  80  0.0008x
0.0012x  20  0.0008x
0.002x  20
20
x
0.002
x  10, 000
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Supply = Demand
 To find the price p p  60  0.0012x
back-substitute
 60  0.0012 10, 000 
x = 10,000 into
 72
the Supply Eqn
5. State: The
equilibrium point
is (10,000, 72).
•
The Graph
serves as
the Chk
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Break Even Analysis
 When a company manufactures x units
of a product, it spends money. This is
total cost and can be thought of as a
function C, where C(x) is the total cost
of producing x units. When a company
sells x units of the product, it takes in
money. This is total revenue and can
be thought of as a function R, where
R(x) is the total revenue from the sale of
x units.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Profit Function
 Total profit is the money taken in less
the money spent, or total
revenue minus total cost.
 Total profit from the production and sale
of x units is a function P given by
Profit = Revenue – Cost
or
P(x) = R(x) – C(x)
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Production Cost

There are two types of costs.
1. Costs which must be paid whether a
product is produced or not, are called
fixed costs.
2. Costs that vary according to the amount
being produced are called variable costs.

The sum of the fixed cost and variable
cost gives the total cost.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Profit

A specialty wallet company has fixed costs that
are $2,400. Each wallet will cost $2 to produce
(variable costs) and will sell for $10.
a) Find the total cost C(x) of producing x wallets.
b) Find the total revenue R(x) from the sale of x wallets.
c) Find the total profit P(x) from the production and sale
of x wallets.
d) What profit will the company realize from the
production and sale of 500 wallets?
e) Graph the total-cost, total-revenue, and total-profit
functions. Determine the break-even point.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Profit

SOLUTION
a) Total cost is given by
•
C(x) = (Fixed costs) plus (Variable costs)
•
C(x) = 2,400 + 2x.
– where x is the number of wallets produced.
b) Total revenue is given by
•
R(x) = 10x
– $10 times the no. of wallets sold
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Profit

SOLUTION
c) Total profit is given by
•
•
•
P(x) = R(x) – C(x)
= 10x – (2,400 + 2x)
= 8x – 2,400.
d) Total profit for 500 units
•
P(500) = 8(500) – 2,400
•
= 4,000 – 2,400
•
Chabot College Mathematics
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= $1,600.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Profit
R(x) = 10x
 The graphs
of R(x), 4,000
C(x),
3,500
and P(x) 3,000
2,500
in $
2,000
Break-even point
C(x) = 2400 + 2x
P(x) = 8x – 2400
1,500
1,000
500
0 50
100 150 200 250 300 350 400 450 500 550
Wallets sold
-2500
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Break-Even Point
 In the Previous Example C x   2 x  2400
note the Cost and
Rx   10 x
Revenue functions
 At BreakEven the Revenues just barely
cover the Costs; i.e., R(xBE) = C(xBE)
 Find BreakEven for the Wallet Factory
CxBE   2 xBE  2400  10 xBE  Rx
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Example  Break-Even Point
 CarryOut
2 xBE  2400  10 xBE
10 xBE  2 xBE  2400
8 xBE  2400
xBE  2400 8  300
 Thus this Wallet production operation
becomes Profitable at production levels
of greater than 300 wallets
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
WhiteBoard Work
 Problems From §3.2 Exercise Set
• 38, 50

Washable
Wallet
Chabot College Mathematics
39
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
All Done for Today
Supply
&
Demand
SeeSaw
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
41
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt