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Chabot Mathematics
§3.2a System
Applications
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Chabot College Mathematics
1
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Review §
3.1
MTH 55
 Any QUESTIONS About
• §’s3.1 → Systems of Linear Equations
 Any QUESTIONS About HomeWork
• §’s3.1 → HW-08
Chabot College Mathematics
2
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
System Methods Compared
 We now have three distinctly different ways to solve
a system. Each method has strengths & weaknesses
Method
Strengths
Weaknesses
Graphical
 Solutions are displayed
visually. Works with any
system that can be graphed.
 Inexact when solutions involve
numbers that are not integers or
are very large and off the graph.
Substitution
 Always yields exact solutions.
 Easy to use when a variable
is alone on one side of an
equation.
 Introduces extensive
computations with fractions
when solving more complicated
systems.
 Solutions are not graphically
displayed.
Elimination
 Always yields exact solutions.  Solutions are not graphically
 Easy to use when fractions or displayed.
decimals appear in the system.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solving Application Problems
1. Read the problem as many times as
needed to understand it thoroughly.
Pay close attention to the questions
asked to help identify the quantity the
variable(s) should represent. In other
Words, FAMILIARIZE yourself with the
intent of the problem
•
Often times performing a GUESS &
CHECK operation facilitates this
Familiarization step
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Solving Application Problems
2. Assign a variable or variables to
represent the quantity you are looking
for, and, when necessary, express all
other unknown quantities in terms of
this variable. That is, Use at LET
statement to clearly state the
MEANING of all variables
•
Frequently, it is helpful to draw a
diagram to illustrate the problem or to set
up a table to organize the information
Chabot College Mathematics
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Bruce Mayer, PE
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Solving Application Problems
3. Write an equation or equations that
describe(s) the situation. That is,
TRANSLATE the words into
mathematical Equations
4. Solve the equation(s); i.e., CARRY
OUT the mathematical operations to
solve for the assigned Variables
5. CHECK the answer against the
description of the original problem (not
just the equation solved in step 4)
Chabot College Mathematics
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Bruce Mayer, PE
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Solving Application Problems
6. Answer the question
asked in the
problem. That is,
make at
STATEMENT in
words that clearly
addressed the
original question
posed in the
problem description
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Problem Solving

Two angles are supplementary. One
of the angles is 20° larger than three
times the other. Find the two angles
1. Familarize
•
Recall that two angles are supplementary
if the sum of their measures is 180°.
We could try and guess, but instead let’s
make a drawing and translate.
Let x and y represent the measures
of the two angles
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Problem Solving
1. Familarize
with Diagram
x
y
2. Translate

Since the angles are supplementary, one
equation is x + y = 180 (1)

The second sentence can be rephrased
and translated as follows:
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Problem Solving
2. Translating

Rewording and Translating
One angle is 20 more than three times the other
y
•
We now have a
system of two
equations and
two unknowns.
Chabot College Mathematics
10
=
20 + 3x
(2)
x  y  180
1
2
y  20  3x
Bruce Mayer, PE
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Example  Problem Solving
3. Carry
Out
x  y  180
1
2
y  20  3x
x  20  3x  180
4x  20  180
4 x  160  x  40
•
Sub x = 40°
in Eqn-1
x  y  180
40 y  180
y  140
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Problem Solving
4. Check
•
If one angle is 40° and the other is 140°,
then the sum of the measures is 180°.
Thus the angles are supplementary.
If 20° is added to three times the smaller
angle, we have 3(40°) + 20° = 140°,
which is the measure of the other angle.
The numbers check.
5. STATE
•
One angle measures 40°
and the other measures 140°
Chabot College Mathematics
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Bruce Mayer, PE
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Elimination Applications
 Total-Value Problems
 Mixture Problems
Chabot College Mathematics
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Bruce Mayer, PE
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Total Value Problems
 EXAMPLE: Lupe sells concessions
at a local sporting event.
• In one hour, she sells 72 drinks.
The drink sizes are
– small, which sells for $2 each
– large, which sells for $3 each.
• If her total sales revenue was $190, how
many of each size did she sell?
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Drinks Sold
1. Familiarize.

•
Suppose (i.e., GUESS) that of the 72
drinks, 20 where small and 52 were large.
•
The 72 drinks would then amount to a
total of 20($2) + 52($3) = $196.
Although our guess is incorrect (but
close), checking the guess has
familiarized us with the problem.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Drinks Sold
1. Familiarize – LET:
•
s = the number of small drinks and
•
l = the number of large drinks
2. Translate.
•
Since a total of 72 drinks were sold, we
must have s + l = 72.
•
To find a second equation, we reword
some information and focus on the
income from the drinks sold
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Drinks Sold
2. Translate.
•
$3
Translating & Rewording
Income from
Income from
small drinks Plus large drinks
$2s

Thus we have
Constructed
the System
Chabot College Mathematics
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+
$2
Totals
$190
=
$190
$3l
s  l  72
2s  3l  190
1
2
Bruce Mayer, PE
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Example  Drinks Sold
3. Carry Out s  l  72 1
l  72  s
3
2s  3l  190
Solve (1) for l
2s  372  s  190
Use (3) to Sub for l in (2)
2s  216  3s  190 Use Distributive Law
216 190  3s  2s Combine Terms
s  26 Simplify to find s

Sub s = 26 in (3) to Find l
l  72  26
Chabot College Mathematics
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l  46
Bruce Mayer, PE
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2
Example  Drinks Sold
4.
Check: If Lupe sold 26 small and 46
large drinks, she would have sold 72
drinks, for a total of:
26($2) + 46($3) = $52 + $138 = $190
5.
State: Lupe sold
•
26 small drinks
•
46 large drinks
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Problem-Solving TIP
 When solving a
problem, see if
it is patterned
or modeled
after a problem
that you have
already solved.
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Problem Solving
 A cookware consultant sells two sizes of
pizza stones. The circular stone sells for
$26 and the rectangular one sells for
$34. In one month she sold 37 stones. If
she made a total of
$1138 from the sale
of the pizza stones,
how many of each
size did she sell?
Pizza Stone
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Pizza Stones
1. Familiarize – When faced with a new
problem, it is often useful to compare it
to a similar problem that you have
already solved. Here instead of $2 and
$3 drinks, we are counting $26 & $34
pizza stones. So LET:
•
c = the no. of
circular stones
•
r = the no. of
rectangular stones
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Pizza Stones cont.2
2. Translate – Since a total of 37 stones
were sold, we have: c + r = 37
•
Tabulating the Data Can be Useful
Circular
Rectangular
Cost per pan
$26
$34
Number of
pans
c
r
26c
34r
Money Paid
Chabot College Mathematics
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Total
37
$1138
c + r = 37
26c + 34r = 1138
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Pizza Stones
2. Translate – We have translated to a
system of equations:
c + r = 37
(1)
26c + 34r = 1138 (2)
3. Carry Out  Multiply Eqn(1) by −26
 26c  26r  962

Add Eqns (2)&(3):
Chabot College Mathematics
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3
26c  34r  1138
 26c  26r  962
0c  8r  176
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
2
3
Example  Pizza Stones cont.4
3. Carry Out – Solve for r
8r  176  r  176 8  r  22
•
Find c using Eqn (1):
c  r  37  c  22  37  c  37  22  15
4. Check: If r = 22 and c = 15, a total of
37 stones were sold. The amount paid
was 22($34) + 15($26) = $1138 
5. State: The consultant sold 15 Circular
and 22 rectangular pizza stones
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Mixture Problem
 A Chemical Engineer wishes to mix
a reagent that is 30% acid and
another reagent that is 50% acid.
 How many liters of each should be
mixed to get 20 L of a solution that
is 35% acid?
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Acid Mixxing
1. Familiarize. Make a drawing and
then make a guess to gain familiarity
with the problem
•
The Diagram
t
liters
+
30% acid
Chabot College Mathematics
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f
liters
=
50% acid
20
liters
35% acid
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Acid Mixxing
 To familiarize ourselves with this
problem, guess that 10 liters of each
are mixed. The resulting mixture will
be the right size but we need to check
the Pure-Acid Content:
0.30(10)  0.50(10)  8 L.
 Our 10L guess produced 8L of
pure-acid in the mix, but we need
0.35(20) = 7L of pure-acid in the mix
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Acid Mixxing
2. Translate:
 LET

•
t = the number of liters of the 30% soln
•
f = the number of liters of the 50% soln
Next Tabulate the calculation of the
amount of pure-acid in each of the
mixture components
Chabot College Mathematics
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Bruce Mayer, PE
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Example  Acid Mixxing
 Pure-Acid Calculation Table
First
Solution
Second
Solution
Mixture
Number of
Liters
t
f
20
Percentage
of Acid
30%
50%
35%
Amount of
Acid
0.30t
0.50f
7
t + f = 20
0.30t + 0.50f = 7
 The Table
t  f  20,
Reveals a System
0.30t  0.50 f  7.
of Equations
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Acid Mixxing
3. Carry Out: Solve Eqn System
t

f
0.30t  0.50 f
 20

7
1
2
 Eliminate f by
multiplying both sides –0.50t – 0.50f = –10
of equation (1) by −0.5 0.30t + 0.50f = 7
and adding them to the
–0.20t = –3
corresponding sides of
t = 15.
equation (2):
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Acid Mixxing
 To find f, we substitute
15 for t in equation (1)
and then solve for f:
15 + f = 20
f=5
 Obtain soln (15, 5), or t = 15 and f = 5
4. Check: Recall t is the of liters of 30%
soln and f is the of liters of 50% soln
Number of liters: t + f = 15 + 5 = 20
Amount of Acid: 0.30t + 0.50f
= 0.30(15) + 0.50(5) = 7
Chabot College Mathematics
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
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Wage Rate
 Ethan and Ian are twins. They have
decided to save all of the money they earn
at their part-time jobs to buy a car to share
at college. One week, Ethan worked 8
hours and Ian worked 14 hours. Together
they saved $256. The next week, Ethan
worked 12 hours and Ian worked 16 hours
and they earned $324.
 How much does each twin make per hour?
• i.e.; What are the Wage RATES
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Wage Rate
 In This Case LET:
• E ≡ Ethan’s Wage Rate ($/hr)
• I ≡ Ian’s Wage Rate ($/hr)
 Translate: Ethan worked 8 hours and
Ian worked 14 hours. Together they
saved $256
8∙{Ethan’s Rt} plus 14∙{Ian’s Rt} is $256
8E  14I  256
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Wage Rate
 Translate: Ethan worked 12 hours
and Ian worked 16 hours and they
earned $324.
12∙{Ethan’s Rt} plus 16∙{Ian’s Rt} is $324
12E  16I  324
 Now have
2-Eqn
System
Chabot College Mathematics
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 14 I
 256
12 E  16 I
 324
8E
Bruce Mayer, PE
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Example  Wage Rate
 Carry Out
8E
12 E
96 E
 14 I
 25612
 324 8
 16 I
 168I
 96 E  128I
40 I
Chabot College Mathematics
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
3072
  2592

480
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Wage Rate
480
 Carry Out 40 I  480  I 
 12
40
 Sub I = 12 into 1st eqn to Find E
8E  1412  256  8E  256  168
 8E  88  E  88 8  E  11
 State Answer
• Ethan Earns $11 per hour
• Ian Earns $12 per hour
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Geometry

The perimeter of a fence around the
children’s section of the community
park is 268 feet. The length is 34 feet
longer than the width. Find the
dimensions of the park.
1. Familiarize: Draw a Diagram and LET:
•
l ≡ the length
•
w ≡ the width
Chabot College Mathematics
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l
w
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Geometry
2. Translate.
 The Perimeter is 2l + 2w.
The perimeter is 268 feet.
2l + 2w = 268
 The length is 34 ft more than the width
l
Chabot College Mathematics
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= 34
+
w
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Geometry

Now have a system of two equations
and two unknowns.
•
2l + 2w = 268
(1)
2l  2w  268
•
l = 34 + w
(2)
l  34  w
3. Solve for w
using
Substitution
Method
Chabot College Mathematics
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2(34  w)  2w  268
68  2w  2w  268
68  4w  268
4w  200
w  50
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Geometry
 Sub 50 for w in one of the Orignal Eqns
l = 34 + w = 34 + 50 = 84 feet
4. Check: If the length is 84 and the
width is 50, then the length is 34 feet
more than the width, and the perimeter
is: 2(84) + 2(50), or 268 feet
5. State: The width is 50 feet and the
length is 84 feet
Chabot College Mathematics
41
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Mixture Problem
 A coffee shop is considering a new
mixture of coffee beans. It will be
created with Italian Roast beans costing
$9.95 per pound and the Venezuelan
Blend beans costing $11.25 per pound.
The types will be mixed to form a 60-lb
batch that sells for $10.50 per pound.
 How many pounds of each type
of bean should go into the blend?
Chabot College Mathematics
42
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Coffee Beans
1. Familiarize – This problem is similar
to one of the previous examples.
•
•
•

Instead of pizza stones we have coffee beans
We have two different prices per pound.
Instead of knowing the total amount paid, we
know the weight and price per pound of the
new blend being made.
LET:
•
•
i ≡ no. lbs of Italian roast and
v ≡ no. lbs of Venezuelan blend
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Coffee Beans
2. Translate – Since a 60-lb batch is
being made, we have i + v = 60.
•
Present the information in a table.
Italian
Number of
pounds
v
60
Price per
pound
$9.95 $11.25
Value of
beans
9.95i
Chabot College Mathematics
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i
Venezuelan New Blend
11.25v
i + v = 60
$10.50
630
9.95i + 11.25v = 630
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Coffee Beans cont.4
2. Translate - We have translated
1
i  v  60
to a system
of equations 9.95i  11.25v  630 2
3. Carry Out - When equation (1) is solved
for v, we have: v = 60 − i.
•
We then substitute for v in equation (2).
9.95i  11.2560  i  630
9.95i  675 11.25i  630
1.30i  45  i  45 1.3  i  34.61
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Coffee Beans cont.5
3. Carry Out - Find v using v = 60 − i.
v  60  i  60  34.61  v  25.39
4. Check - If 34.6 lb of Italian Roast and
25.4 lb of Venezuelan Blend are
mixed, a 60-lb blend will result.
•
The value of 34.6 lb of Italian beans is
34.6•($9.95), or $344.27.
•
The value of 25.4 lb of Venezuelan Blend
is 25.4•($11.25), or $285.75,
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Coffee Beans cont.6
4. Check – cont.
•
so the value of the blend is
[$344.27 + $285.75] = $630.02.
•
A 60-lb blend priced at $10.50 a pound is
also worth $630, so our answer checks
5. State – The blend should be made from
•
•
34.6 pounds of Italian Roast beans
25.4 pounds of Venezuelan Blend beans
Chabot College Mathematics
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Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Simple Interest
 If a principal of P dollars is borrowed for
a period of t years with interest rate r
(expressed as a decimal) computed
yearly, then the total interest paid at the
end of t years is
I  P  r t
 Interest computed with this formula is
called simple interest. When interest is
computed yearly, the rate r is called an
annual interest rate
Chabot College Mathematics
48
Bruce Mayer, PE
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Example  Simple Interest
 Ms. Jeung invests a total of $10,000 in
Bonds from blue-chip and technology
Companies. At the end of a year, the
blue-chips returned 12% and the
technology stocks returned 8% on the
original investments.
 How much was invested in each type of
Bond if the total interest earned was
$1060?
Chabot College Mathematics
49
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest



Familiarize:
We are asked to find two amounts:
•
that invested in blue-chip Bonds
•
that invested in technology Bonds.
If we know how much was invested in
blue-chip Bonds, then we know that
the rest of the $10,000 was invested in
technology Bonds.
Chabot College Mathematics
50
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest


TRANSLATE: LET
•
x ≡ amount invested in blue-chip bonds.
•
y ≡ amount invested in tech bonds
Tabulate Interest Income
Invest P
t
I = Prt
Blue
x
0.12
1
0.12x
Tech
y
0.08
1
0.08y
Chabot College Mathematics
51
r
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest
 Translate TOTAL INTEREST statement
Interest from
Interest from
Total
+ technology =
Blue-chip
Interest
0.12 x  0.08 y  $1060
 Translate TOTAL Principal statement
{BluChip Prin.} plus {Tech Prin.} is $10k
x  y  $10000
Chabot College Mathematics
52
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest
 Carry Out: Solve Principal Eqn for y
and then sub into Interest Eqn
y  10,000  x & 0.12 x  0.08 y  1060
0.12x  0.08 10, 000  x   1060
0.12x  0.08 y  1060100 Clear Decimals 
12x  8 10, 000  x   106, 000
12x  80, 000  8x  106, 000
4x  26, 000
Chabot College Mathematics
53
x  6500
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest
 Back Substitute x  6500 & y  10,000  x
x = 6500 into
 y  10000  6500  3500
Principal Eqn
y

3500
to find y
 Check
$3500  $6500  $10,000

12% of $6500  $780
8% of $3500  $280
Total interest earned = $1060.
Chabot College Mathematics
54

Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Example  Simple Interest
 State: Ms. Jeung invested
$3500 in technology Bonds and
$6500 in blue-chip Bonds.
Chabot College Mathematics
55
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
WhiteBoard Work
 Problems From §3.2 Exercise Set
• 22

Peppermint
Patty
Puzzled
Chabot College Mathematics
56
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
All Done for Today
Commercial
Coffee Bean
Blending
Chabot College Mathematics
57
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
–
Chabot College Mathematics
58
Bruce Mayer, PE
[email protected] • MTH55_Lec-10_sec_3-1_2Var_LinSys_ppt