1. Linear Equations and Linear Inequalities
Chapter 1 Phase 1
Linear Equations and Linear Inequalities

2. Equations Statement where two expressions are equal Examples x + 2 = 9
x = 5x – 3
11x = 5x + 6x

3. 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  = 5  5 = 5

4. 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

5. Linear Equations First Degree Equations Example
Highest exponent on the x variable is a 1
When there is not an exponent shown, the default exponent is 1
Example
x = x1
Phase 1 DB Part A

6. Solving Linear Equations – Starting Simple
Solve the following linear equation for x
x + 5 = 8
==========
x = 3
Check it by plugging in x = 3 into the original equation
3 + 5 = 8
8 = 8
Subtract 5 from both sides of the equation

7. Solving Linear Equations – Little Harder
Solve the following linear equation for x
5x + 4 = x
-3x x
============
2x = 12
2x = 8
/ /2
x = 4
Collect “like terms”
Subtract 3x on both sides of the equation
Collect “like terms”
Subtract 4 on both sides of the equation
MML Homework Question #1
“Isolate x” by dividing both sides by 2
Reminder: Forward slash represents division

8. Solving Linear Equations – Little Harder
Solve the following linear equation for x (CHECK IT!)
5x + 4 = x
5(4) + 4 = (4) =
24 = 24
We just found x = 4 PLUG IT IN!
MML Homework Question #1

9. 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 –
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
MML Homework Chapter1 Question #5

10. 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 =
21 = 21

11. Solving Linear Equations – Horrid Fractions
Solve the following linear equation for x
Find common denominator
How about 15? Multiply every fraction by 15!
MML Homework Question #4

12. Solving Linear Equations – Horrid Fractions

13. Solving Linear Equations – Horrid Fractions
CHECK IT! Plug in x = 5
Common Denominator = 15

14. Solving for a Particular Variable
Solve the following equation for y
6x – 2y = 4
- 6x x
===========
- 2y = 4 – 6x
/(-2) /(-2)
y = x
Subtract both sides by 6x
Divide both sides by NEGATIVE 2
NOTE: You are dividing every term on the right side by -2
MML Homework Question #6

15. 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
============
F – 32 = (9/5) C
* (5/9) *(5/9)
=============
(5/9)(F-32) = C
Subtract both sides by 32
Multiply both sides by (5/9)
MML Homework Question #7

16. Linear Inequalities Inequality Symbols < Less Than
> Greater Than
<= Less Than or Equal
>= Great Than or Equal
Phase 1 DB Part A

17. 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

18. Solving Linear Inequalities
Solve the following inequality
Subtract 3 on both sides of the inequality
Divide negative value of -2 to both sides (sign reverses)
MML Homework Chapter1 Question #1

19. Double Inequalities Variable is between two values
Example: a < x <b
Which means a < x
and x < b
x is between a and b

20. 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)
MML Homework Chapter1 Question #3

21. 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!

22. 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

23. 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

24. 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 x
=============
5 = 0.25x
/ /0.25
==============
= 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

25. 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?
Phase 1 DB

26. Solving Word Problems with Inequalities
Solve the inequality for x
Multiply 5 to all sides of the inequality

27. Solving Word Problems with Inequalities
Add the current grades and subtract value from all sides
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?

28. 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

29. Rectangular / Cartesian Coordinate System
x-axis
abscissa
Origin
y-axis ordinate

30. Where would we graph the following points?
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

31. Graphing Linear Equations vs. Linear Inequalities
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
Phase 1 DB

32. 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)

33. 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
If A = 0 and B ≠ 0 graphs a horizontal line
If A ≠ 0 and B = 0 graphs a vertical line
MML Homework Chapter1 Question #14

34. 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)
MML Homework Chapter1 Question #16

35. Slope of a Line
Find the slope of a line that passes through the points (-2, 3) and (4, -5)
MML Homework Chapter1 Question #16

36. 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
MML Homework Question #10

37. 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 – = 3x – 6 + 4
y = 3x – 2
MML Homework Question #15

38. 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)
4/22/2017
© Cindy Roberts 2007

39. x-intercept (5,0) y-intercept (0,5) 4/22/2017
MML Homework Question #13
4/22/2017

40. Distance Formula Find the distance between the two points
Given two points: P1(x1,y1) and P(x2,y2)
Find the distance between the two points
Phase 1 DB Part B #1

41. Distance Formula
Find the distance between (3,2) and (7,4)

42. 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)

43. Application of Linear Equation
Find Slope from points (0,300) and (100,4300)

44. 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

45. Application of Linear Equation
Cost Equation
Slope = 40
Given the point (0,300) gives the y intercept of 300
Cost = 40x + 300

46. Graphing in Excel x Cost = 40x + 300 =40*A2+300 50 =40*A3+300 100
=40*A2+300 50
=40*A3+300
100
=40*A4+300
150
=40*A5+300
200
=40*A6+300 x
Cost = 40x + 300
300 50
2300
100
4300
150
6300
200
8300

47. Graph

48. Break—Even Point
Equilibrium point
Supply = Demand
Cost = Revenue

49. 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
7800
2.37

50. 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)
Year
Supply (mil bu)
Demand (mil bu)
Price ($/bu)
1990
7500
7900
2.28
1991
7800
2.37

51. Application – Supply and Demand
Use point-slope form to find the supply equation
p – p1 = m(x – x1)
p – 2.28 = (x – 7500)
p – 2.28 = x –
p = x

52. 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)
Year
Supply (mil bu)
Demand (mil bu)
Price ($/bu)
1990
7500
7900
2.28
1991
7800
2.37

53. Application – Supply and Demand
Use point-slope form to find the demand equation
p – p1 = m(x – x1)
p – 2.28 = (x – 7900)
p – 2.28 = x
p = x

54. Application – Supply and Demand
Find the equilibrium point set supply = demand
Supply: p = x
Demand: p = x
x = x
Add x on both sides
x x = x x
x = 9.39
Subtract on both sides
x – = 9.39 –
x =
Divide on both sides
x / = /
x = 7820

55. Application – Supply and Demand
Plug in x = 7820 into either equation
p = x
p = (7820)
p = 2.352
Equilibrium Point (7820, 2.352)

56. 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

57. 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)
$146
$112 1
$150
$120 2
$156
$125 3
$160
$128 4
$164
$130 5
$174
$136
Phase 1 IP
MML Homework Chapter1 Question #19

58. Graph Data in Excel (Scatterplot)

59. Find the Linear Regression Equation
Select Graph
Chart Tools / Layout / Trendline / More Trendline Options
Trend/Regression Type = Linear
Checkbox Display Equation on chart

60. Linear Regression Equation

61. Linear Regression Equation
y = x
Slope =
Every year the average price in thousands goes up
Y-intercept =
At year = 0 or year = 2000 the “predicted price” is $ in thousands

62. Predicting Predict the average price in the year 2010
2010 would be 10 years after 2000 so x = 10
y = x
y = (10)
y =
y = or $198,193 is the predicted average price for the year 2010