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Chapter 1
Phase 1
Linear Equations and Linear Inequalities
Equations
 Statement where two expressions are equal
 Examples
x+2=9
 x = 5x – 3
 11x = 5x + 6x
Solutions to Equations
 To solve an equation means to find all numbers that make the
equation a true statement
 These numbers are called the “solution”
 Equation have “equal signs”
 What is on the left of the equal sign must have the same value as
what is on the right of the equal sign
 Example
 x = 4 for the equation x + 1 = 5 because if we substitute 4 into
the equation for x it is a true statement
x+1=54+1=55=5
Properties of Equations
 Equivalent equations (Equal) are a result of
 Same quantity is added or subtracted on both sides of the equal sign
 Same nonzero quantity is multiplied or divided on both sides of the
equal sign
Linear Equations
 First Degree Equations
 Highest exponent on the x variable is a 1
 When there is not an exponent shown, the default exponent is 1
 Example
 x = x1
Solving Linear Equations – Starting Simple
 Solve the following linear equation for x
x+5=8
- 5 -5
==========
x
=3
Subtract 5 from both
sides of the equation
 Check it by plugging in x = 3 into the original equation
x+5=8
3+5=8
8 =8
Solving Linear Equations – Little Harder
 Solve the following linear equation for x
Collect “like terms”
5x + 4 = 12 + 3x
Subtract 3x on both
-3x
- 3x
sides of the equation
============
Collect “like terms”
2x + 4 = 12
Subtract 4 on both
-4
-4
sides of the equation
============
“Isolate x” by
2x
= 8
dividing both sides
/2
/2
by 2
============
Reminder: Forward slash represents division
x
=4
Solving Linear Equations – Little Harder
 Solve the following linear equation for x (CHECK IT!)
5x + 4 = 12 + 3x
5(4) + 4 = 12 + 3(4)
20 + 4 = 12 + 12
24 = 24
We just found x = 4
PLUG IT IN!
Solving Linear Equations – Wow!
 Solve and check the following linear equation
 5x – 2(x – 3) = 2(x – 2) + 15
 Remove parentheses
 5x – 2x + 6 = 2x – 4 + 15
 Remember a negative and a negative make a positive
 Collect like terms on each side
 3x + 6 = 2x + 11
 Subtract 6 from both sides
 3x + 6 – 6 = 2x + 11 – 6
 3x = 2x + 5
 Subtract 2x from both sides
 3x – 2x = 2x + 5 – 2x
x=5
CHECK IT!
 Verify results by plugging in x = 5
 5x – 2(x – 3) = 2(x – 2) + 15
 5(5) – 2(5 – 3) = 2(5 – 2) + 15
 Remember Order of Operations, PEMDAS
 25 – 2(2) = 2(3) + 15
 25 – 4 = 6 + 15
 21 = 21
Solving Linear Equations – Horrid Fractions
 Solve the following linear equation for x
x 2 1
 
5 3 3
 Find common denominator
 How about 15? Multiply every fraction by 15!
 x
 2
1
15   15   15 
5
 3
 3
Solving Linear Equations – Horrid Fractions
 x
 2
1
15   15   15 
5
 3
 3
3x  10  5
3x  10  10  5  10
3x  15
3 x 15

3
3
x5
Solving Linear Equations – Horrid Fractions
 CHECK IT! Plug in x = 5
x 2 1
 
5 3 3
5 2 1
 
5 3 3
15 10 5
 
15 15 15
Common Denominator = 15
5 5

15 15
Solving for a Particular Variable
 Solve the following equation for y
6x – 2y = 4
- 6x
-6x
===========
- 2y = 4 – 6x
/(-2)
/(-2)
===========
y = -2 + 3x
Subtract both sides
by 6x
Divide both sides by
NEGATIVE 2
NOTE: You are
dividing every term on
the right side by -2
Solving for a Particular Variable
 The following equation can be used for converting
temperatures from Fahrenheit (F) to Celsius (C).
 Solve the equation for C
F = (9/5) C + 32
Subtract both sides
by 32
- 32
- 32
============
Multiply both sides
F – 32 = (9/5) C
by (5/9)
* (5/9) *(5/9)
=============
(5/9)(F-32) = C
Linear Inequalities
 Inequality Symbols
 < Less Than
 > Greater Than
 <= Less Than or Equal
 >= Great Than or Equal
Solving Linear Inequalities
 Properties
 Inequality will stay the same if
 Add or Subtract a real value
 Multiply or divide a positive value
 Inequality will reverse if
 Multiply or divide a negative value
 NOTE: Multiplication by 0 is not allowed
Solving Linear Inequalities
 Solve the following inequality
 2 x  3  4
 Subtract 3 on both sides of the inequality
 2 x  3  3  4  3
 2 x  7
 Divide negative value of -2 to both sides (sign reverses)
 2x  7

2 2
7
x
2
Double Inequalities
 Variable is between two values
 Example:
a < x <b
 Which means
a<x
and
x<b
 x is between a and b
Solving Double Inequalities
 Solve:
-9 < 2x + 1 < 7
 Looking for all numbers x such that 2x + 1 is between -9 and 7
 Isolate x in the middle
 -9 < 2x + 1 <7
 Subtract 1
 -9 – 1 < 2x + 1 – 1 < 7 – 1
 -10 < 2x < 6
 Divide 2
 -10/2 < 2x/2 < 6/2
 -5 <x < 3 or (-5,3)
Solving Word Problems
 Understand what the question is asking you to find
 That becomes the variable
 Use only needed information to solve problem
 See if answer makes sense!
Solving Word Problems with Equations
 “Equilibrium point” or “Break-Even point”
 Where supply = demand; sales = costs, etc.
 Set up the equations
 Set the equations equal to each other
 Solve for the desired variable
Solving Word Problems with Equations
 A company produces fasteners that cost two cents per
fastener to make plus five dollars for fixed costs. The
company sells the fasteners for twenty-seven cents each.
Where is the break-even point?
 Set up equations
 Let x = the number of fasteners produced
 Cost = 0.02x + 5
 Revenue = 0.27x
 Set Cost = Revenue and solve for x
Solving Word Problems with Equations
 Set Cost = Revenue and solve for x
 Cost = 0.02x + 5
 Revenue = 0.27x
0.02x + 5 = 0.27x
-0.02x
-0.02x
=============
5 = 0.25x
/0.25 /0.25
==============
20 = x
Move x’s to one side by
subtracting .02x on both
sides of equation
Divide both sides by
0.25x to isolate x
Company must sell 20
fasteners to break even
Solving Word Problems with Inequalities
 In your Math class you have completed four tests with grades
of 68, 82, 87 and 89. If all tests are weighted equally, what
grade range should you earn on the last test to ensure a B
(80-89 range) in the class?
 How do you find an average?
68  82  87  89  x
80 
 89
5
Solving Word Problems with Inequalities
 Solve the inequality for x
 Multiply 5 to all sides of the inequality
68  82  87  89  x
80 
 89
5
 68  82  87  89  x 
80 * 5  
 * 5  89 * 5
5


400  68  82  87  89  x  445
Solving Word Problems with Inequalities
 Add the current grades and subtract value from all sides
400  68  82  87  89  x  445
400  326  326  x  326  445  326
74   x  119
 Need to make at least a 74 on the last test
 Did we really need the right side of the inequality and does the
value make sense?
Interval Notation
Interval Notation Inequality Notation
[a,b]
a≤x≤b
[a,b)
a≤x<b
(a,b]
a<x≤b
(a,b)
a<x<b
(-∞,a]
x≤a
(-∞,a)
x<a
[b,∞)
x≥b
(b,∞)
x>b
Rectangular / Cartesian Coordinate System
x-axis
abscissa
Origin
y-axis
ordinate
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Plotting Points
Where would we graph the following points?
(-5,1), (0,3), (4,3), (3,0), (3,-3), (0,-4), (-5,-2)
(0,3)
(4,3)
(-5,1)
(3,0)
(-5,-2)
(3,-3)
(0,-4)
Plotting (x,y) coordinates
Graphing Linear Equations vs. Linear Inequalities
 Linear Equations
 Graph of a line
 Ordered pairs (x,y) solutions
to the equation
 Linear Inequalities
 Graph of an “area”
(line could be included)
 Ordered pairs (x,y) solutions
to the inequality
Linear Equation in Two Variables
 Standard Form
 Ax + By = C
 Example
 3x – 2y = 6
 One solution x = 4 and y = 3 or the ordered pair (4,3)
 Find other solutions by picking a value for one variable and
solving for the other
 Let x = 2
 3(2) – 2y = 6  6 – 2y = 6  -2y = 0  y = 0
 Thus, another solution is (2,0)
Graph Linear Equation in Two Variables
 Ax + By = C is a line
 If A ≠ 0 and B ≠ 0, then the equation can be rewritten as
A
C
y x
B
B


C
If A = 0 and B ≠ 0 graphs a horizontal line y 
B
A
If A ≠ 0 and B = 0 graphs a vertical line
y
B
Slope of a Line
 Numerical measurement of the steepness of a line
 Line passes through two distinct points
 P1(x1,y1) and P(x2,y2)
y2  y1
vertical change (rise)
m

x2  x1 horizontal change (run)
Slope of a Line
 Find the slope of a line that passes through the points (-2, 3)
and (4, -5)
y2  y1  5  3  8
4
m



x2  x1 4  (2) 6
3
Other Forms of Linear Equations
 Slope-Intercept Form
 y = mx + b
 m is the slope (steepness of the line)
 b is the y intercept (where the line crosses the y axis)
 Example
y = 3x + 5
 Where 3 is the slope and 5 is the y-intercept
Other Forms of Linear Equations
 Point-Slope Form
 y – y1 = m(x – x1)
 Example:
 Line has slope of 3 and point (2,4)
 y – 4 = 3(x – 2)
 Can now solve for y to put in slope-intercept form
y – 4 = 3x – 6
y – 4 + 4 = 3x – 6 + 4
y = 3x – 2
Intercepts
 Where the line crosses the axis
 x-intercept is where the line crosses the x-axis or where y=0
 Example: (5,0)
 y-intercept is where the line crosses the y-axis or where x=0
 Example: (0,3)
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© Cindy Roberts 2007
5/24/2017
x-intercept (5,0)
y-intercept (0,5)
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5/24/2017
Distance Formula
 Given two points: P1(x1,y1) and P(x2,y2)
 Find the distance between the two points
d  ( x2  x1 )  ( y 2  y1 )
2
2
Distance Formula
 Find the distance between (3,2) and (7,4)
d  ( x2  x1 )  ( y 2  y1 )
2
d  (7  3)  (4  2)
2
d  (4)  (2)
d  16  4
d  20  4.47
2
2
2
2
Application of Linear Equation
 Cost Equation
 The management of a company that manufacturers roller
skates have fixed costs (costs at 0 output) of $300 per day and
total costs of $4,300 per day at an output of 10 pairs of skates
per day. Assume that cost is C is linearly related to output x
 Find the slope of the line joining the points with outputs of 0
and 100; that is, the line passing through (0,300) and
(100,4300)
Application of Linear Equation
 Find Slope from points (0,300) and (100,4300)
y2  y1 4300  300 4000
m


 40
x2  x1
100  0
100
Application of Linear Equation
 The management of a company that manufacturers roller
skates have fixed costs (costs at 0 output) of $300 per day and
total costs of $4,300 per day at an output of 10 pairs of skates
per day. Assume that cost is C is linearly related to output x
 Find an equation of the line relating output to cost. Write the
final answer in the form Cost = mx + b. Then graph the
equation from 0 ≤ x ≤ 200
Application of Linear Equation
 Cost Equation
 Slope = 40
 Given the point (0,300) gives the y intercept of 300
 Cost = 40x + 300
Graphing in Excel
x
0
50
100
150
200
x
0
50
100
150
200
Cost = 40x + 300
=40*A2+300
=40*A3+300
=40*A4+300
=40*A5+300
=40*A6+300
Cost = 40x + 300
300
2300
4300
6300
8300
Graph
Cost Function
Cost per day (dollars)
$165
Cost = 40x + 300
$160
$155
$150
$145
0
1
2
3
Output per day
4
5
Break—Even Point
 Equilibrium point
 Supply = Demand
 Cost = Revenue
Application – Supply and Demand
 The following table lists the supply and demand for barley in
the United States during two recent years
 Assume the relationship between supply and price is linear
and between demand and price is linear
 Find the supply and demand equations
 Find the equilibrium point
Year
Supply (mil bu) Demand (mil bu) Price ($/bu)
1990
7500
7900
2.28
1991
7900
7800
2.37
Application – Supply and Demand
 Find the supply equation of the form p = mx + b, where p is
the price per bushel in dollars and x is the corresponding
supply in billions of bushels
 Use two points to find the slope (7500, 2.28) and (7900, 2.37)
y2  y1 2.37  2.28 0.09
m


 0.000225
x2  x1 7900  7500 400
Year
Supply (mil bu) Demand (mil bu) Price ($/bu)
1990
7500
7900
2.28
1991
7900
7800
2.37
Application – Supply and Demand
 Use point-slope form to find the supply equation
p – p1 = m(x – x1)
p – 2.28 = 0.000225(x – 7500)
p – 2.28 = 0.000225x – 1.6875
p = 0.000225x + 0.5925
Application – Supply and Demand
 Find the demand equation of the form p = mx + b, where p
is the price per bushel in dollars and x is the corresponding
demand in billions of bushels
 Use two points to find the slope (7500, 2.28) and (7900, 2.37)
y2  y1 2.37  2.28
0.09
m


 0.0009
x2  x1 7800  7900  100
Year
Supply (mil bu) Demand (mil bu) Price ($/bu)
1990
7500
7900
2.28
1991
7900
7800
2.37
Application – Supply and Demand
 Use point-slope form to find the demand equation
p – p1 = m(x – x1)
p – 2.28 = -0.0009(x – 7900)
p – 2.28 = -0.0009x + 7.11
p = -0.0009x + 9.39
Application – Supply and Demand
 Find the equilibrium point set supply = demand
 Supply: p = 0.000225x + 0.5925
 Demand: p = -0.0009x + 9.39
 0.000225x + 0.5925 = -0.009x + 9.39
 Add 0.0009x on both sides
 0.000225x + 0.5925 + 0.0009x = -0.009x + 9.39 + 0.0009x
 0.001125x + 0.5925 = 9.39
 Subtract 0.5925 on both sides
 0.001125x + 0.5925 – 0.5925 = 9.39 – 0.5925
 0.001125x = 8.7975
 Divide 0.001125 on both sides
 0.001125x / 0.001125 = 8.7975 / 0.001125
 x = 7820
Application – Supply and Demand
 Plug in x = 7820 into either equation
 p = 0.000225x + 0.5925
 p = 0.000225(7820) + 0.5925
 p = 2.352
 Equilibrium Point (7820, 2.352)
Mathematical Model
 Construct a mathematical model to represent data
 Solve the mathematical model
 Use the model for predicting unknown values
 Model using Excel’s Add Trendline Function
 Linear Regression
 “Best Fit” Line that models data
Linear Regression Example
 The following table contains recent average and median
purchase prices for a house in Texas.
 Model the data in Excel to “find the regression equation”
 Let x = # of years since 2000
Year Average Price (in thousands) Median Price (in thousands)
0
$146
$112
1
$150
$120
2
$156
$125
3
$160
$128
4
$164
$130
5
$174
$136
Graph Data in Excel (Scatterplot)
Average price in thousands
Texas Real Estate Prices
$180
$175
$170
$165
$160
$155
$150
$145
$140
0
1
2
3
Year
4
5
6
Find the Linear Regression Equation
 Select Graph
 Chart Tools / Layout / Trendline / More Trendline Options
 Trend/Regression Type = Linear
 Checkbox Display Equation on chart
Linear Regression Equation
Average price in thousands
Texas Real Estate Prices
$180
$175
$170
$165
$160
$155
$150
$145
$140
y = 5.3143x + 145.05
0
1
2
3
Year
4
5
6
Linear Regression Equation
 y = 5.3143x + 145.05
 Slope = 5.3143
 Every year the average price in thousands goes up 5.3143
 Y-intercept = 145.05
 At year = 0 or year = 2000 the “predicted price” is $145.05 in thousands
Predicting
 Predict the average price in the year 2010
 2010 would be 10 years after 2000 so x = 10
 y = 5.3143x + 145.05
 y = 5.3143(10) + 145.05
 y = 53.143 + 145.05
 y = 198.193 or $198,193 is the predicted average price for
the year 2010