Download 1 - University of Wisconsin System

Curricular Redesign & Emerging Technology Grant
Identification
Project title: Curricular Redesign Grant Proposal
2010-2011
Intermediate Algebra Video Project
Principal Investigator Theresa Adsit
(campus affiliation): University of Wisconsin Green Bay
Other significant persons: None.
Start and end dates of the June 2010 to May 2011
project:
Date of this report: August 26, 2011
See also: Instructional Development Council at the University of Wisconsin
Green Bay
Abstract
The purpose of the Intermediate Algebra Video Project was to enhance student learning in
Intermediate Algebra by moving the bulk of instructional lecture material to videos which the
student was assigned to view outside of class and thus free up valuable class time to engage in
active learning techniques emphasizing real life examples and a deeper understanding of
algebra.
Project Narrative
Implementation:
Once the videos were completed and students watched these outside of class, during class we
had time for a hands on learning activity each week, including:
• Direct variation demonstrated by a pendulum
• Using manipulatives to further understanding of cubic volumes and square areas i.e.
why doesn’t a square foot = 12 square inches?
• simplifying versus solving, using a scale to demonstrate the difference
• Constructing parabolas, understanding why they are used for manufacturing satellite
dishes and flash lights
• Budgeting with a spread sheet in the general access computer lab
• Using linearization to demonstrate a real life optimization example
• Using a graphing calculator to enhance understanding of the quadratic formula and
quadratic and rational inequalities
• Proving the Pythagorean Theorem with squares and triangles then examining
applications
• What does slope mean in real life
• Simplifying real formulas and real calculus examples
• What are irrational numbers and how do you approximate their values
Videos:
The Intermediate Algebra Video Collection consists of 98 videos averaging 9 minutes each.
Students watch about 65 minutes of video each week. I ended up needing to create a
workbook for the students to work in along with me as they watch the videos. The workbook
allowed me to record many of the notes that the student would have had to have written
otherwise, thus speeding up students’ viewing time. Students seem to like the workbook and
use it as an abbreviated text. This Workbook is Appendix A. The students can work together
with me as they watch the videos, focusing on the problems we are working on. These were
recorded using the UWGB Mediasite recording studio managed by the Adult Degree Program.
These may be viewed at:
http://uwgb.mediasite.com/mediasite/Catalog/pages/catalog.aspx?catalogId=067c0bdb-189847cd-a798-78cfbbdbaac8
Exceptions to the original proposal
The proposal was carried out as written except that I ended up needing to write an Algebra
Workbook to accompany the videos.
Conclusion
In general the students did not do any better or worse. Giving them a very similar final exam to
those given in other years resulted in similar grades. Since their grades were already good,
improvement in grades was not the main goal.
My primary goal was for students to understand the importance and relevance of mathematics.
This was tested by looking at the student evaluation question regarding importance and
relevance of the class. Unfortunately, this evaluation question did not show any change in
students’ perspective. My feeling is that too few students participated in the hands on
activities. In retrospect, making these activities required as some part of the grade would have
been a better idea.
Students who did attend hands on activities seemed to really enjoy having real mathematically
significant activities. Even some applications which I thought would bore them were met with
interest and enthusiasm. Since I surveyed the students after each application, it was easy to
see what they enjoyed most. It appears that the more an example can be used for something
they would like to accomplish, the better they like it.
After surveying students midway through the semester, I found that 75% either were OK with
or very pleased with having a hybrid course. Some students really liked the flexibility of having
lectures on video. About 25% did not like having lectures on videos. Several students
resentfully expressed that they had to take this hybrid format because all of the other
traditional lecture sections were filled. In retrospect, teaching three out of the six offered
sections of math 101 as a hybrid course did not leave students enough choice of course format.
It was disappointing that so few students who were doing poorly actually took advantage of the
extra time I had available to help them in class. It was unusual for more than 5 students out of
a class of 30 to get extra help during the question and answer portion of class.
From a teacher’s point of view, having time to teach real world applications and more time to
spend answering individual questions has been wonderful. Additionally, I also had classroom
time to be able to include a weekly exam which helped the students by providing immediate
feedback in how they were doing.
While I was disappointed in not seeing improved understanding of the relevance of
mathematics, it was encouraging to see the students’ enjoyment of hands on projects.
Additionally, I am excited to now have available video lectures for students who miss class for
various reasons. Also, I will continue to use the Algebra Workbook in class with students.
Appendix A
UW-Green Bay
College Algebra Workbook
Theresa Adsit
To the Student:
You can be successful at learning to do algebra! Even if Algebra is not your area
of expertise, hard work and perseverance can make up for a lack of natural
aptitude. The key is to carefully do your homework problems and seek help if
you are not able to find a correct answer. Repetition is vital to remembering
how to work out the different types of problems. Work enough problems so that
you know how to do each type and then work enough more so that you can do
them quickly.
This workbook is designed to be used in conjunction with Saxon’s text, Algebra
2; either the second or third editions will have corresponding lesson numbers.
We will not be covering every lesson in the text, only those which I teach at UW
Green Bay for Intermediate Algebra.
Warm Regards,
Theresa Adsit
Lesson 2 Exponent Rules
Definitions:
Def: an = a•a•…•a
n times
Def: a0 = 1, if a≠0
𝟏
𝟏
Def: a-n=𝒂𝒏 , 𝒂−𝒏 = an, if a≠0
Properties:
1)
aman = am+n
2)
𝒂𝒎
3)
(am)n = amn
4)
(ab)m = ambm
𝒂𝒏
= am-n
ex. 23•22 =
𝟐𝟑
ex. 𝟐𝟓 =
ex. (23)2 =
Ex. Simplify:
𝒏𝟎 𝒑−𝟐 (𝒂𝒏𝟐 )−𝟑
𝟐𝒑𝒑𝟐 (𝒂−𝟑 )𝟎 𝒂−𝟐
factors to the 0 power = 1
multiply outer exponent
move factors with negative exp
multiply by adding exp
divide by subtracting exp
Ex. Simplify:
𝟑𝒃−𝟐 𝒒𝒄𝟑 (𝒃−𝟑 )−𝟐
(𝟐𝒒𝟎 𝒃)𝟑 (𝒄−𝟑 )
Ex. Simplify:
(𝒏𝟐 )−𝟐 𝒌𝒎𝟎 𝒏𝟑
(𝟐𝟎 𝒌−𝟐 𝒎)𝟐 𝒏−𝟑 𝒏
Lesson 3 Evaluating Formulas, Adding Like Terms
Evaluating:
Ex. Evaluate a2b – ab2 if a = -3 and b = 2
Ex. Evaluate (ab – a2b) – b if a = 2 and b = -5
Adding Like Terms:
Ex. Simplify by combining like terms:
𝟑 −𝟐
𝒂𝒌 𝒑
𝟐𝒂−𝟏 𝒑𝟐 𝒌
𝟑𝒑−𝟏 𝒌
𝟓𝒑−𝟐 𝒑𝟐 𝒂−𝟏 𝒌−𝟑
+
− −𝟏 −𝟐 +
𝒌𝟒
𝒂 𝒑𝒌
𝒑−𝟐
simplify each term
combine coefficients of like terms
(those w/ same variables, exp.)
Ex. Simplify by combining like terms:
𝟓𝒃𝟑 𝒒−𝟏 𝒎𝟐
𝒎𝟑
−
𝟐𝟎 𝒎
𝒃−𝟐
−
𝟐𝒎−𝟐 𝒒
𝒎−𝟏 𝒒𝟐 𝒃−𝟑
Lesson 4: The Distributive Property, Solving Equations
The Distributive Property: c(a + b) = ca + cb
ex. 5(2 + 3) = 5(2) + 5(3)
ex. Expand:
𝟓𝒑𝒒𝟎 𝟐𝒑−𝟐 𝒎−𝟑 𝒎−𝟐 𝒒𝟑
�
−
�
𝒎−𝟐
𝒒𝟐
𝒑𝟑
distribute
simplify each term
Ex. Expand:
𝟐𝒚𝒙−𝟏 𝟑𝟎 𝒚𝟎 𝒌𝒌−𝟏 𝒙𝟐 𝒙𝒌
�
−
�
𝒌𝟐
𝒙
𝒚𝒚𝟑
Solving Equations:
Ex. Solve:
𝟑
𝟐𝟓𝒙 +
𝟓
=
𝟑
𝟏𝟓
𝟐
𝒙−𝟐
write as improper fractions
multiply both sides of the equation by
the LCD, canceling denominators
continue to solve
Ex. Solve:
𝟐
𝟏
𝟓
− �𝟒 − 𝒙� = 𝒙 + 𝟏
𝟑
𝟐
𝟔
Lesson 5 Introduction to Word Problems
Ex. The sum of a number and 5 is multiplied by 7. The result is 10 less than 2
times the number. What is the number?
Ex. Two times a number is added to 5 and the sum is multiplied by 4. The result
is 12 more than 6 times the number. What is the number?
Ex. Two fifths of the calculus students preferred blue notebooks. If 120
students were in calculus, how many did not prefer blue notebooks?
Lesson 6 Problems with Decimal Numbers,
Consecutive Integer Problems,
Problems with Decimals:
Ex. Solve: 0.02x - 0.13 = 0.6x + 3.1
Multiply both sides by
100
Solve
Ex. Luella found that a honey bee perched on 0.013 of the Dandelions examined.
If 10,000 Dandelions were examined, how many did not have a honey bee?
Consecutive Integer Problems:
Integers: …-3, -2, -1, 0, 1, 2, 3…
Consecutive Odd Integers: …-5, -3, -1, 1, 3, 5…
Consecutive Even Integers: …-6, -4, -2, 0, 2, 4, 6…
For variables:
Consecutive Integers: x, x + 1, x + 2, …
Consecutive Even/Odd Integers: x, x + 2, x + 4,…
Ex. Find three consecutive odd integers such that two times the sum of the first
and the third is 17 less than 5 times the second.
Ex. Find three consecutive integers such that three times the sum of the second
and third is four less than five times the second.
Lesson 7: Percentage Word Problems
Recall: 60% = 0.60, 13% = 0.13, 1.5% = .015
Ex.Twenty-four is eighty percent of what number?
Ex.Fourty-nine hundred is what percent of seven hundred?
Lesson 8 Graphing Linear Equations
Ex. Graph 2x + 3y = 6
Slopes of Lines: m =
x y
𝒓𝒊𝒔𝒆
𝒓𝒖𝒏
=+
𝒖𝒑
𝒓𝒊𝒈𝒉𝒕
=−
𝒅𝒐𝒘𝒏
𝒓𝒊𝒈𝒉𝒕
Slope intercept form: y = mx + b
m = slope, b = y intercept
Ex. Graph 3x – 2y = 10
Lesson 9 More Percent Word Problems
Ex. Seventy percent of the Algebra students handed in their homework on
Monday. If sixty Algebra students did not hand in their homework on Monday,
how many Algebra students were there?
Ex. The Beekeeper was astounded because this year’s honey production
increased by 150%. If she harvested 750 pounds of honey this year, how many
pounds did she harvest last year?
Lesson 10 Distance and the Pythagorean Theorem
Recall the Pythagorean Theorem:
In a right triangle with legs of length a and b and hypotenuse of length c,
c
b
a
a2 + b2 = c2
Ex. Find a:
9
11
a
Ex. Use the Pythagorean Theorem to find the distance between (-3, 2) and (5, -1).
Lesson 11 Addition of Rational Expressions
Fractions must have the same denominator in order to be added or subtracted.
𝟏
Ex. Add 𝟏𝟐 +
𝟑
𝟏𝟎
Find the LCD
Use the
Numerator/Denominator rule to get
the LCD
Add the numerators over
the LCD
𝟏
Ex. Add 𝟐𝟒 +
𝟏
𝟏𝟖
Ex. Add
𝒙
+
𝒂𝒌
𝒃𝒎
+ 𝒚
+ 𝟑+
𝟐𝒄
𝒄𝟐 𝒅 𝟑
𝟐𝒃𝟐
Ex. Add
𝒂𝒃
𝒂𝟐
Ex. Add
𝒋𝒙
𝟓𝒌𝒏𝟑
+
𝒎
+
𝒌𝒏
𝒚
𝒏𝟐
Lesson 12:Finding the Equation of a Line From its Graph
Recall
Slopes of Lines: m =
𝒓𝒊𝒔𝒆
𝒓𝒖𝒏
=+
𝒖𝒑
𝒓𝒊𝒈𝒉𝒕
Slope intercept form: y = mx + b
=−
𝒅𝒐𝒘𝒏
𝒓𝒊𝒈𝒉𝒕
m = slope, b = y intercept
Ex. Find the equation of the line below:
Ex. Find the equation of the line below:
Ex. Find the equation of the line below:
Ex. Find the equation of the line below:
Lesson 13 Solving Systems of Equations
Using Substitution
Ex. Use substitution to solve:
x = 2y – 3
3x + 5y = 24
Ex. Use substitution to solve:
3x + 2y = 15
5x + y = -10
Lesson 14: Finding the Equation of a Line
Ex. Find the equation of the line passing through the points
(2, -3) and (-1, 7)
Graph the line
Calculate slope from graph
Insert (x,y) and m into
y = mx + b and use algebra
to find b.
Ex. Find the equation of the line passing through
(-2, 6) with slope -2/5.
Lesson 15: Solving Systems of Equations Using Elimination:
Ex. Solve: -3x + y = 11
3x – 2y = 2
Add the equations to
eliminate x, Solve
Find x
Ex. Use elimination to solve:
3x – 2y = 22
5x + 3y = 5
Multiply each equation so that the y (or x) coefficients are opposites.
Then add the equations to eliminate a variable and solve.
3x – 2y = 22
5x + 3y = 5
Ex. Solve using elimination:
-2x + 5y = 19
3x + 4y = 6
Lesson 16 Multiplication and Division of Polynomials
Multiplication:
Ex. Multiply: (3x + 2)(5x2 – x + 4)
Division:
First consider long division of numbers:
232 ÷ 11 =
1)
2)
3)
4)
5)
𝟐𝟑𝟐
𝟏𝟏
������
= 𝟏𝟏|𝟐𝟑𝟐
Divide to choose what to multiply by (first terms)
Multiply (both terms)
Subtract
Bring Down
Repeat
Ex. Divide 3 – 2x + 5x2 + 2x3 by -2 + x
Check:
(x – 2)(2x2 + 9x + 16) + 35
2x3 + 9x2 + 16x – 4x2 – 18x – 32 + 35
2x3 + 5x2 - 2x + 3
Ex. Divide x3 + 2x + 7 by -2 + x
Ex. Divide -2 + 3x – 4x2 – 2x3 by -1 + x
Lesson 17: Subscripted Variables
Ex. Solve: RBTB + RGTG = 240, RB = 40, RG = 60, TB + TG = 8
Ex. Solve: RJTJ + RCTC = 1000, RJ = 60, RC = 50, TJ + TC = 16
Lesson 18 Ratio Word Problems
Ex. At the kennel club, the ratio of mixed breed dogs to pure bred dogs was 7 to
3. If there were 120 dogs at the kennel club, how many were pure bred?
Given:
Unstated:
The question regards which two quantities?
Ex. It took 2 cups of chocolate chips to mix 8 cups of cookie dough. How many
cups of other ingredients are required to mix 28 cups of cookie dough?
Given:
Unstated:
The question regards which two quantities?
Ex. It took 300 kg of lead to make 5000 kg of the new compound. How many kg
of other materials are required to make 12000 kg of the new compound?
Lesson 19: Word Problems With Items of Value
Ex. There were thirty-three dimes and quarters in all whose value was six
dollars and thirty cents. How many were dimes and how many were quarters?
Number of items:
Value of items:
Ex. Adult tickets cost twenty dollars each while children’s tickets cost only
fifteen dollars each. If Bruce bought a total of fifteen tickets and spent $260.00,
how many children’s tickets did he buy?
Lesson 20: Introduction to Radicals, Parallel Lines
Radicals:
Consider:
√𝟒 = 𝟐 because 2•2 = 4
Also √𝟐 ≈ 𝟏. 𝟒𝟏𝟒
√𝟐√𝟐 = 2
However, 1.414 • 1.414 = 1.999396
Similarly, √𝟓 ≈ 𝟐. 𝟐𝟑𝟔
√𝟓√𝟓 = 𝟓
However, 2.236 • 2.236 = 4.999696
A square root equals that number which multiplied times itself equals the
radicand.
We will only consider the principal, or positive, root.
Ex. Simplify: √𝟓𝟎𝟎
find the prime factorization
of 500
a square root times itself
equals the radicand
Ex. Simplify √𝟕𝟐
Ex. Simplify:
𝟑√𝟏𝟓𝟎 − 𝟓√𝟐𝟏𝟔
simplify each term
only like terms may be
added or subtracted
Ex. Simplify 𝟐√𝟑(𝟑√𝟑 − 𝟓√𝟏𝟐)
distribute,
coefficients multiply, radicands
multiply
simplify each term
Ex. Simplify 𝟐√𝟑 ∙ 𝟒√𝟏𝟖 − 𝟔√𝟐𝟎𝟎
Parallel Lines:
The slopes of parallel lines are equal.
Ex. Find the equation of a line parallel to 2x – 3y = 6 which passes through the
point (-1, 2).
solve the given equation
for y, find its slope
fill in m, (x,y) into y = mx + b
and use algebra to find b
Ex. Find the equation of the line parallel to 3x – 2y = 12 which passes through (2, 4)
Lesson 21: Scientific Notation, Word Problems With Two Equations and Two
Variables
Scientific Notation:
𝟏
𝟏
Very small numbers: 0.00034 = 3.4 x 10-4 = 3.4 x 𝟏𝟎𝟒 = 3.4 x𝟏𝟎,𝟎𝟎𝟎
Very large numbers: 3,200,000 = 3.2 x 106 = 3.2 x 1,000,000
A number in proper scientific notation has one digit to the left of the decimal.
Ex. Simplify:
𝟐𝟎𝟎𝟎 × 𝟏𝟎−𝟐 𝟎. 𝟎𝟎𝟎𝟏𝟖 × 𝟏𝟎𝟑
∙
𝟗𝟎𝟎𝟎
𝟎. 𝟐𝟎𝟎𝟎𝟎
write each number in
proper scientific notation
make a separate fraction
out of the powers of ten
use exponent rules to
simplify the fraction
Word Problems With Two Variables:
Ex. The ratio of two numbers is 3 to 7. Their sum is 120. What are the two
numbers?
Ex. The sum of two numbers is 81. Their difference is 35. What are the two
numbers?
Lesson 22: Introduction to Uniform Motion Problems
Recall: D = R∙T
for example, 600 miles = 60 mph ∙10 hr.
For this lessons problems, two distances will be equal.
Identify Distance, Rate and Time with capital D, R and T. Use subscripts to
indicate which individual the distance, rate or time is associated with. For
example, distance for Julie: DJ
Ex. Victoria made the trip to Grandmother’s in 10 hours. It took her brother
Michael just 8 hours to make the same trip because his speedometer was
broken. If Victoria was traveling at 50 mph, how fast was Michael traveling?
draw arrows to represent
distances traveled
make an equation relating
distances
replace each distance with
the equivalent rate ∙ time
fill in the known values
solve
Ex. On the way to Springfield, Jon’s Piper flew at just 200 mph into the wind but
on the way back he caught a tail wind and traveled at 250 mph. If the total
traveling time was 9 hours, how far was it to Springfield?
Lesson 23: Solving Systems of Equations by Graphing
Ex. Find the solution by graphing:
2x + y = 3
y = -3x + 6
graph each line
the intersection is the solution
check your answer
Lesson 24: Equations With Fractions
Ex. Solve:
𝟑𝒙−𝟏
𝟒
−
𝟓
𝟔
=
𝟏
𝟑
use prime factorization
to find LCD
multiply both sides
of the equation by LCD,
canceling denominators
solve
Ex. Solve:
𝟐𝒙
𝟓
−
𝟔+𝟑𝒙
𝟏𝟎
=𝟒
Lesson 25 Factoring the GCF, Cancellation
Factoring the GCF:
Ex. Factor:
16x3yz2 – 8x2y3z3 + 4xyz2
Cancellation:
Do not cancel part of a sum or a difference. Sums and differences must
completely cancel or not cancel at all.
Ex.Simplify:
(𝟓)(𝟑𝟒)
(𝟓)
can cancel a
factor in a product
Ex. Simplify:
Ex. Simplify
𝒙+𝟓
𝟓
can’t cancel part
of a sum
𝟓𝒙+𝟓
𝟓
factor and then cancel
Ex. Simplify:
𝟏𝟐𝒚𝟑 − 𝟒𝒚
Ex. Simplify:
𝟐𝟕𝒂𝟒 − 𝟏𝟐𝒂𝟐
𝟒𝒚
𝟑𝒂𝟐
Lesson 26: Factoring Trinomials x2 + bx + c
Recall: Multiplication:
(x + 3)(x – 2)
F: Firsts
O: Outers
I: Inners
L: Lasts
Thus, factoring requires two numbers whose product is the constant and whose
sum is the x coefficient.
Ex. Factor: x2 + 3x – 10
product = -10, sum = 3?
Ex. Factor: -x2 – x + 56
factor out -1
(-1 remains in final answer)
product = -56, sum = 1?
Ex. Factor: -10x2 + x3 + 16x
factor off GCF,
write in descending order
Lesson 27 Rational Expressions Whose Denominators Contain Sums
Recall: Rational Expressions must have the same denominators before they can
be added or subtracted.
Check denominators to see if they factor.
A sum in a denominator is a distinct factor in the LCD.
Ex. Add
𝒙+𝟐
𝒙−𝟏
+ 𝟑−
𝟒
𝒙𝟐
Find LCD
Each fraction needs to have
the LCD
Combine numerators
over LCD
Ex. Add:
𝟑
𝒂+𝟐
−
𝟒
𝒂
Ex. Add:
𝒎−𝟏
𝒎𝟐 + 𝒎−𝟔
+
𝟑
𝒎(𝒎−𝟐)
Lesson 28: Complex Fractions, Introduction to Rationalizing Denominators
Complex Fractions:
Recall:
𝟑
𝟓
𝟏
𝟕
=
top fraction.
Ex. Simplify:
𝟑
𝟓
∙
𝒙
𝒚
𝒙+𝟑
𝒚
𝟕
𝟏
=
𝟐𝟏
𝟓
Invert (flip) and multiply the bottom fraction times the
Ex. Simplify:
𝒂
𝒃−𝒄
𝒅
𝒃−𝒄
Introduction to Rationalizing Denominators:
Do not leave radicals (roots) in the denominator.
Multiply the numerator and the denominator by the square root in the
denominator.
Ex. Simplify:
𝟔
𝟓√𝟑
𝟔
Notice that 𝟓√𝟑 ≈ . 𝟔𝟗𝟐𝟖 and
Ex. Simplify:
10
7√2
𝟐√𝟑
𝟓
≈ . 𝟔𝟗𝟐𝟖
Lesson 29: Uniform Motion Problems in Which the Sum of Two Distances is
Known
Recall:
At first:
D1 = D 2
R1T1 = R2T2
Now we will also see,
D1 + D 2 = k
R1T1 + R2T2 = k
Ex. Bruce and Wendy parted at O’Hare International Airport when Wendy’s
plane left at 9:00 am. flying East at 350 mph. Bruce’s plane sped West at 300
mph but did not leave until 11:00 am. What time was it when the two were 2000
miles apart?
draw arrows representing distance
distances add to 2000
replace each distance
with corresponding rate x time
fill in known values
make a 2nd equation regarding times
solve
answer the question
Ex. Scarlett rode in the wagon at 5 mph for part of the 40 mile trip to Tara until
her horse became worn out and she had to walk the remaining distance at
3mph. If the total trip took 10 hours, how far did she walk?
Lesson 31: Perpendicular Lines
Perpendicular lines form a right angle.
The slopes of perpendicular lines are opposite reciprocals:
ex.
𝟏
-2, 𝟐
𝟑
𝟓
𝟓
,−𝟑
𝟕 𝟒
−𝟒,𝟕
Ex. Find the equation of the line perpendicular to the line 2x – 3y = 12 passing
through the point (-2, 1).
solve for y and find the
slope of the given line
find the perpendicular slope
fill m, (x, y) into y = mx + b
and solve for b
write the equation of theline
Ex. Find the equation of the line perpendicular to the line 3x – 4y = 8 passing
through the point
(1, -3).
Lesson 32: Combining Roots Containing Fractions
𝟑
𝟕
Ex. Simplify: �𝟕 + �𝟑
make individual roots in the
numerator& denominator
rationalize the denominator
need LCD in both terms
add like terms
Ex. Simplify:
𝟑
𝟐
𝟑�𝟐 − 𝟒�𝟑
Ex. Simplify:
𝟐
𝟓
𝟐�𝟓 − 𝟑�𝟐
Lesson 33: Complex Fractions Containing Sums
First we combine fractions to attain a single fraction in the numerator and a
single fraction in the denominator. Then invert and multiply the denominator.
Ex. Simplify:
𝒙
𝟐
+
𝒚 𝟑𝒚
𝟐
𝒚
“little fractions” in the numerator need to have the
same LCD to add
add “little fractions”
invert and multiply denominator
reduce (cancel) if possible,
remember, don’t cancel part of a sum
Ex. Simplify:
𝟑𝒂𝒃
+𝟐
𝒄
𝟏 𝒃
+
𝒄 𝟑𝒄
Ex. Simplify:
𝒋
𝟏
+
𝒌+𝟑 𝒌
𝟓
𝒋+𝒏
Lesson 34: Uniform Motion Problems in Which One Distance is Shorter than the
Other
First,
D1 = D 2
Then,
R1T1 = R2T2
D1 + D2 = k
R1T1 + R2T2 = k
Now we will also see,
D1 + k = D2
R1T1 + k = R2T2
Ex. Jack and Jill raced up the hill. In 50 seconds, Jill reached the top. She
turned just in time to see Jack fall down only 100 m from the top. If Jill ran at 8
meters per second, how fast was Jack running before he fell?
Ex. Jack had a 45 mile head start and was traveling at 15 mph when Jill set out
to catch him. If Jill was traveling twice as fast as Jack, how long did it take for
her to catch Jack?
Lesson 35: Rational Exponents
Recall: √𝟐√𝟐 = 𝟐 because √𝟐 is that positive number which when multiplied
times itself yields 2.
𝟑
By √𝒙 we mean that number which when multiplied times itself 3 times yields x.
𝟑
Ex.√𝟖= 2 because 2∙2∙2 = 8
𝟑
𝟑
𝟑
𝟑
𝟑
𝟑
Ex.√𝟐 √𝟐 √𝟐 = 𝟐
𝟑
Notice, √𝟐 ≈ 1.414 but √𝟐 ≈ 1.260
Ex.√𝟓 √𝟓 √𝟓 = 𝟓
Radicals may be written using fractional exponents.
𝟐
Ex.√𝟑 = √𝟑 = 𝟑𝟏⁄𝟐
𝟑
Ex.√𝟓 = 𝟓𝟏⁄𝟑
𝒏
In general, √𝒂𝒎 = ( √𝒂)𝒎 = 𝒂𝒎⁄𝒏
𝒏
Ex. Simplify 𝟐𝟕−𝟏⁄𝟑
negative exponents move
factors to the other side of
the fraction
write as a cubed root
simplify the radical
Ex. Simplify: 𝟏𝟐𝟓𝟐⁄𝟑
Ex. Simplify: −𝟔𝟒−𝟐⁄𝟑
Lesson 36: Multiplication and Division of Rational Expressions
Recall:
𝟓
𝟑
𝟕
𝟐
=
𝟓
𝟑
÷
𝟕
𝟐
=
𝟓
𝟑
∙
𝟐
𝟕
=
𝟏𝟎
𝟐𝟏
Divide rational expressions by inverting and multiplying.
Ex. Simplify:
𝑥 2 − 6𝑥 + 5
𝑥 2 − 2𝑥 − 15
÷ 2
𝑥2 + 𝑥 − 2
𝑥 + 10𝑥 + 21
invert and multiply
factor and cancel
Ex. Simplify:
𝒙𝟑 − 𝟒𝒙𝟐 − 𝟐𝟏𝒙
𝒙𝟑 − 𝟖𝒙𝟐 + 𝟕𝒙
÷
𝒙𝟐 + 𝟓𝒙 + 𝟔
𝒙 + 𝟐
Lesson 38: Powers of Sums, Solving by Factoring
Powers of Sums:
Ex. Expand: (x + 3)3
Solving by Factoring:
Consider:
If 8∙n = 0, then n = 0.
If m∙n = 0, then m = 0 or n = 0 or both
The Zero Factor Principle:
If k∙p∙…∙q = 0
then k = 0 or p = 0…or q = 0
Ex. Solve:
x3 = 3x2 + 28x
need 0 on one side of
the equation
factor
set each factor = 0 and solve
Ex. Solve: 0 = 162 + 3x2 – 45x
Lesson 39: Factoring the Difference of Two Squares
Consider: Conjugates - the same two terms with the opposite middle sign
ex:
2x + 3, 2x – 3
-x + 5, -x – 5
4x – 1, 4x + 1
Consider the product of conjugates:
(2x + 3)(2x – 3) = 4x2 + 6x – 6x – 9 = 4x2 – 9
(-x + 5)(-x – 5) = x2 + 5x – 5x – 25 = x2 – 25
(4x – 1)(4x + 1) = 16x2 + 4x – 4x – 1 = 16x2 – 1
Thus the difference of two squares can be factored as the product of two
conjugates.
Ex. Solve by factoring:
49x2 – 36 = 0
factor
the difference of squares
set each factor = 0
solve
Ex. Solve by factoring:
81x2 – 121 = 0
Lesson 40: Solving an Abstract Equation
Ex. Solve for b:
𝟐+
𝟑
𝒙
= 𝒚𝒛 +
𝒂
𝒃
multiply both sides of the
equation by the LCD to cancel
the denominators
isolate “b” terms on one side
factor “b” off
divide out “b’s” coefficient
Ex. Solve for f:
𝒂
𝒃
− 𝟐=
𝒆
𝒇
𝒄
+𝟓
Ex. Solve for m:
𝒙
𝒎
+ 𝒄=
𝒚
𝒂
Lesson 41: Converting Units
Unit examples: lbs., ft., mi., m, sec.
A unit multiplier is a fraction with equivalent values in the numerator and
denominator but different units.
Some Unit Multipliers:
𝟏 𝒇𝒕
𝟏𝟐 𝒊𝒏
𝟓𝟐𝟖𝟎 𝒇𝒕
𝟏 𝒎𝒊
Equal
Equal
𝟏 𝒉𝒓
𝟔𝟎 𝒎𝒊𝒏
𝟏 𝒚𝒅
𝟑 𝒇𝒕
Equal
Equal
Each unit multiplier may be written inverted.
We multiply quantities by unit multipliers to cancel units we do not want and
introduce units we do want.
Ex. Use unit multipliers to convert 35 miles to inches.
Consider square inches, cubic inches:
1 in.2
1 in
1 in.
1 in.
3
1 in.
1 in.
Ex.Convert 2 cubic miles to cubic inches.
Ex. Convert 3200 square inches to square feet.
1 in.
Lesson 45: Introduction to Solving Quadratic Equations
Consider: Solve: x2 = 25
x = 5 or x = -5
Ex. Solve:
x2 = 13
take the square root of both
sides of the equation, don’t
forget the ±on the right
Ex. Solve:
𝟏
(x + 𝟑)2 = 12
take the square root of both
sides of the equation, don’t
forget the ±on the right
solve for x
simplify
Ex. Solve:
(x + 4)2 = 15
Lesson 46: More on Simplifying Radicals Involving Fractions, More on
Simplifying Radicals Using Fractional Exponents
Ex. Simplify:
𝟑
𝟕
𝟓� + 𝟐� − 𝟑√𝟖𝟒
𝟕
𝟑
make individual roots in the
numerator& denominator
rationalize the
denominators
simplify the √𝟖𝟒
need LCD in each term
add like terms
Ex. Simplify:
𝟐
𝟓
� − 𝟑� − 𝟒√𝟐𝟓𝟎
𝟓
𝟐
Recall:
𝒏
√𝒂𝒎 = 𝒂𝒎⁄𝒏
Ex. Simplify:
3
5
√𝑎5 𝑏 2 √𝑎3 𝑏
write each factor with rational
exponents
multiply factors with the same
base by adding their exponents
exponents need the same LCD
to be added
Ex. Simplify:
�5√5
write the inner √𝟓 as 51/2
multiply powers of 5 by
adding exponents
express the outer √
½
using an exponent of
multiply the outer
exponent times the inner exponent
Ex. Simplify:
𝟑
�𝟑√𝟑
Lesson 47: More Roots Containing Roots, Converting Rates
Ex. Simplify:
�25√5
write 25 as 52
write the inner √𝟓 as 51/2
multiply powers of 5 by
adding exponents
express the outer√ using an
exponent of ½
multiply the outer exponent times the
inner exponent
Ex. Simplify:
3
�9 4√3
When converting rates, they may be expressed as fractions.
Such as 60 mph =
𝟔𝟎 𝒎𝒊𝒍𝒆𝒔
𝟏 𝒉𝒐𝒖𝒓
The “per” is the fraction bar.
To convert a rate, simply use unit multipliers to convert the time component (if
necessary) and then multiply by more unit multipliers to convert the length
component (if necessary).
Ex. Convert 60 miles per hour to inches per minute.
Ex. Convert 10 miles per second to feet per hour.
Lesson 48: An Introduction to Equations Containing Radicals
Ex. Solve:
√𝒎 + 𝟑 − 𝟓 = −𝟐
isolate the radical
square both sides of
the equation
solve
a check is required
Note: You cannot square an equation in “pieces.”
ex.
2 + 3 = 5 is true but
22 + 32≠ 52
4 + 9 ≠ 25
Ex. Solve:
√𝒙 + 𝟏𝟎 + 𝟑 = −𝟏𝟐
Ex. Solve:
√𝒙𝟐 − 𝟒𝒙 − 𝟖 + 𝟒 = 𝒙
Ex. Solve:
√𝒂 + 𝟏𝟎 + 𝟐 = −𝟐
Lesson 49: Finding the y Intercept of a Line
Ex. Find the equation of the line:
find two points on the line
use them to find the line’s
slope
use one point and m in
y = mx + b to calculate b
Ex. Find the equation of the line.
Lesson 50: Completing the Square to Solve Quadratic Equations, an
Introduction
Recall:
Solve: x2 – 5x + 6 = 0
(x – 3)(x – 2) = 0
x – 3 = 0 or x – 2 = 0
x=3
or x = 2
Also, Solve: (x – 3)2 = 11
�(𝒙 − 𝟑)𝟐 = ±√𝟏𝟏
𝒙 − 𝟑 = ±√𝟏𝟏
𝒙 = 𝟑 ± √𝟏𝟏
Completing the square enables us to solve non-factorable quadratic equations
of the first form by putting them into the second form.
Ex. Solve by completing the square:
x2 + 6x - 8 = 0
need the constant on the other side
write out binomials (x + k)(x + k) (where k is
½ times the x coefficient)
and add k2 to the right
take the square root of both sides
solve for x
Ex. Solve:
x2 = 7x + 9
Ex. Solve:
-6x – 12 + x2 = 0
Lesson 51: Non-real Numbers
Consider: √𝟒 = 2 because 2∙2 = 4
What should √−𝟒 equal?
-2(-2) ≠ -4
2(2) ≠ -4
0(0) ≠ -4
There is no real number whose square = -4.
Def: √−𝟏 = 𝒊
𝑖 2 = −1
Ex.√−𝟒 = √𝟒√−𝟏 = 𝟐𝒊
Ex.√−𝟗 = √𝟗√−𝟏 = 𝟑𝒊
Ex.√−𝟏𝟓 = √𝟏𝟓√−𝟏 = √𝟏𝟓𝒊
Any number that can be written in the form a + bi where a and b are real
numbers is a complex number.
Ex. Simplify:
3iii + 5ii - 7i – 2 + √−𝟐
write√−𝟐 as √𝟐𝒊 , each i2 = -1
combine like terms
Ex. Simplify:
5i3 – 2i5 + 4i – 8 + √−𝟒
Ex. Simplify:
-5i4 + 3iii – 6 + 3i - √−𝟗
Lesson 52: Chemical Mixture Problems In Which the Final Quantity is Known
M
C
final volume
M+C
= Final Volume
%M + %C = %∙Final Volume
Ex. Christy needs 600 ml of 70% witch hazel. She has only a 60% witch hazel
solution on hand and the drug store only sells a 90% witch hazel mixture. How
many ml of 60% witch hazel and 90% witch hazel should Christy mix?
Ex. The chef needs to mix up 12 cups of a 50% brine solution. He has only 40%
brine solution and 80% brine solution on hand. How much of each should he
combine?
Lesson 53: Converting Metric Units
Recall:
2.54 cm = 1 in.
100 cm = 1 m
1000 m = 1 km
Ex. Convert 2.3 square kilometers to square inches.
Ex. Convert 3 cubic meters to cubic feet.
Lesson 55: Advanced Abstract Equations, Word Problems Containing Quadratic
Equations
Ex. Solve for b:
𝟐+
𝟑
𝒙
= 𝒚𝒛 +
𝒂+𝒚
𝒃
multiply both sides of the
equation by the LCD to cancel
the denominators
distribute ( )
isolate “b” terms on one side
factor “b” off
divide out “b’s” coefficient
Ex. Solve for m:
𝒂
𝒎
+
𝟐
𝒙+𝒚
=𝒌
Ex. Find e:
𝒄
𝒙+𝒚
−
𝟓
𝒆
=𝒋
Word Problems Containing Quadratic Equations:
Ex. Find three consecutive integers such that the product of the first and the
third is 23 more than ten times the opposite of the second.
Ex. Find 3 consecutive even integers such that the product of the first times the
second is 10 more than 5 times the opposite of the sum of the second and third.
Lesson 58: Completing the Square With an x2 Coefficient Other Than 1
Ex. Solve by completing the square:
3x2 + 6x - 8 = 0
divide both sides of the equation by 3
need the constant on the other side
write out binomials (x + k)(x + k) (where k is
½ times the x coefficient)
and add k2 to the right
take the square root of both sides
solve for x
Ex. Solve by completing the square:
6 - 4x2 = 7x
Ex. Solve by completing the square:
10x2 = -6x + 3
Lesson 59: Solving Systems of Equations Containing Fractions or Decimals
Multiply each equation so that it no longer contains fractions or decimals. Then
solve using substitution or elimination.
Ex. Solve:
𝟏
𝟓
𝒙+
𝟏
𝟐
𝒚=
𝟕
𝟓
0.01x - 0.04y = -0.19
Ex. Solve:
5x – 0.7y = 3.6
𝟏
−𝟒𝒙 +
𝟏
𝟑
𝒚=
𝟓
𝟏𝟐
Lesson 60: Introduction to Variation
A varies directly as B means that A = kB
thus as A increases, B increases, also as A decreases, B decreases
𝟏
A varies inversely as B means that A = k𝑩
thus as A increases, B decreases, also as A decreases, B increases
k is a number, called the constant of proportionality.
Ex.The pressure of an ideal gas varied inversely with its volume. If four liters of
the gas had a pressure of 100 newtons per square meter, what would be the
pressure of five liters of the gas?
Ex. The mother’s consumption of chocolates varied directly as the number of
children with her in the car. If 2 children resulted in her eating 14 chocolates,
how many chocolates would she eat if 5 children were in the car?
Ex. The pressure of an ideal gas varied inversely with its volume. If 48 liters of
the gas had a pressure of 7 atmospheres, what would be the pressure of 56
liters of the gas?
Lesson 61: Chemical Mixture Problems in Which the Initial Volume is Known
Initial Volume ± X =
(Initial Volume ± X)
%Initial Volume ± %X = %(Initial Volume ± X)
% of chemical of interest
Ex. The 200 lbs of 20% nitrogen fertilizer mix was too rich for the soil. How
much of the 7% nitrogen fertilizer mix should be added to achieve a mix
containing 15% nitrogen?
Ex. Jim made 32 cups of lemonade. It was too sour as it contained only 4%
sugar. How much sugar should Jim add to make the lemonade 25% sugar?
Ex. The 100 gallons of brine solution was not concentrated enough as it
contained 37% water. How many gallons of water should be evaporated until the
mixture contains just 10% water?
Lesson 62: Non-real Solutions When Completing the Square
Ex. Solve by completing the square:
3x2 + 6x + 8 = 0
divide both sides of the equation by 3
need the constant on the other side
write out binomials (x + k)(x + k) (where k is
½ times the x coefficient)
and add k2 to the right
take the square root of both sides
solve for x
Ex. Solve:
2x2 + 10 = 3x
Lesson 64: More Complex Fractions, More Complex Numbers
Complex Fractions:
Ex. Simplify:
𝒄+
𝟐
𝟏
+ 𝟑𝒃
𝒎
need LCD to add
denominator fractions
invert and multiply
need LCD to add C to the
fraction
Ex. Simplify:
𝒎
𝒙
+
𝟓
𝟑+
𝟐𝒃
𝒙
Complex Numbers:
Recall: √𝟐√𝟐 = 𝟐
Also, √−𝟐√−𝟐 = −𝟐
Be careful, √−𝟑√−𝟐 = √𝟑𝒊√𝟐𝒊 = √𝟔𝒊𝟐 = −√𝟔
√−𝟑√−𝟐 ≠ √𝟔
Ex. Simplify:
5iii – 4i4 + 3i – 7 + √−𝟐√−𝟓
write i4 as iiii, write
√−𝟐 = √𝟐𝒊, √−𝟓 = √𝟓𝒊
each i2 = -1
simplify& combine
like terms
Ex. Simplify:
(3i – 2)(4i + 1)
distribute
each i2 = -1
simplify & combine
like terms
Lesson 65: Advanced Substitution
Decide which Rate x Time to substitute into. Completely replace that rate and
time. Distribute. The other Rate x Time appears. Replace that with its numeric
value.
Ex. Solve:
RATA = 90, RBTB = 360, RB = 3RA, TA + TB = 14
Ex. Solve:
RATA = 105, RBTB = 30, RA = 5RB, TA + TB = 17
Lesson 66: LCD Which Contain Opposite Factors
𝟑
Consider: − 𝟕 =
−𝟑
𝟕
𝟑
= −𝟕
Ex. Add:
𝟐
−
𝒎−𝟑
𝒑
need LCD for both
−𝒎+𝟑
rational expressions
multiply the numerator
& denominator of one fraction by -1
add, don’t lose the denominator
Ex. Add
𝟒𝒂−𝟑
−𝒂−𝟐
−
𝒂−𝟓
𝒂+𝟐
Lesson 67: Rationalizing Denominators Containing Sums
Recall:
Conjugates are the same two terms with opposite signs in the middle:
Ex.
2x + 3, 2x – 3
𝟐√𝟓 − 𝟑, 𝟐√𝟓 + 𝟑
-√𝟑 + 𝟏, −√𝟑 − 𝟏
The product of conjugates is special:
(2x + 3)(2x – 3) = 4x2 – 6x + 6x – 9 = 4x2 – 9
(2√𝟓 - 3)(2√𝟓 + 3) = 4∙5 + 6√𝟓 - 6√𝟓 - 9 = 11
(-√𝟑+ 1)(-√𝟑 - 1) = 3 + √𝟑 - √𝟑 - 1 = 2
The product of conjugates is the difference of squares.
Ex. Simplify:
𝟐
−𝟒+ √𝟓
multiply the numerator
and the denominator
by the conjugate of the
denominator
Ex. Simplify:
𝟓
𝟐√𝟓 − 𝟑√𝟔
Ex. Simplify:
1
-2√3 – √7
Lesson 68: Using Your Calculator to Find nth Roots
It is not necessary in this class to have a calculator that can find nth roots.
However, if you think your calculator will do so, this lesson helps you figure out
how.
Method 1:
4
1�
4
Recall: √81=81
Enter in your calculator:
81 ^ (1/4)
The result should be 3
Method 2:
𝒏
Find the √𝒙 key on your calculator. You may need to press a 2nd key first
to access it.
Enter in your calculator:
𝒏
4 √𝒙 81
The result should be 3
If the result is not 3, try:
𝒏
81 √𝒙 4
If none of these methods work and you think your calculator can find nth roots,
please see me after class.
Lesson 70: Abstract Equations With Parenthesis
Ex. Solve for a:
𝒒
𝒓
𝟑
𝟐
= 𝒎 �𝒂+𝒃 − 𝒑�
distribute m
multiply both sides of the
equation by the LCD to cancel
the denominators
distribute ( )
solate “a” terms on one side
factor “a” off
divide out “a’s” coefficient
Ex.Solve for x:
𝒗
+ 𝒚 = 𝒌 �𝟐 +
𝒙
𝒚
�
𝒋+𝟓
Ex. Find k:
𝒋
𝒎
𝒓 = 𝒒 �𝒙−𝒌 + 𝟐 �
Lesson 71: The Quadratic Formula
Derive the quadratic formula by completing the square to solve:
ax2 + bx + c = 0
divide both sides of the equation by a
need the constant on the other side
write out binomials (x + k)(x + k) (where k is
½ times the x coefficient)
and add k2 to the right
need LCD to add fractions
take the square root of both sides
solve for x
Memorize: 𝒙
=
−𝒃 ±�𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
Round and round the mulberry bush
“x is equal to negative b”
The monkey chased the weasel
“Plus or minus the square root”
The monkey stopped to pull up his socks
“of b squared minus 4 a c”
Pop goes the weasel
“All over 2 a”
Ex. Use the quadratic formula to solve:
3x2 – 4 = 5x
Ex. Solve using the quadratic formula:
2x2 - 7x = 3
Lesson 73: Rationalizing the Denominator When the Numerator is a Sum
Ex. Simplify:
𝟑 − 𝟓√𝟐
𝟕√𝟐
multiply the numerator
and the denominator by √𝟐
simplify
Ex. Simplify
𝟓√𝟐− 𝟒√𝟏𝟐
𝟑√𝟐 + 𝟖√𝟑
multiply the numerator and
denominator by the conjugate of the
denominator
simplify each term
combine like terms
reduce if possible, do not cancel part
of a sum
Ex. Simplify
√𝟐𝟒 + 𝟓√𝟐
𝟕√𝟐 − 𝟑√𝟑
Lesson 74: Uniform Motion Problems in Which Both Distances are Known
Ex. Rachel lived farther from her parents than her brother Derek. In fact, she
had to drive 360 miles to get to their parents’ home. Derek decided to bicycle
the 45 miles to his parents’ home. Even though Rachel drove four times as fast
as Derek bicycled, it still took her three hours longer to get home than it took
Derek. What were the rates and times of each of them?
Ex. Brenda was traveling twice as fast as Gwendolyn. That’s why she could
cover 66 miles in just 4 hours more than it took Gwendolyn to cover 21 miles.
What were the rates and times of each?
Lesson 75: Factorable Denominators With Opposite Factors
Ex. Add:
𝒙−𝟒
𝒙+𝟑
−
𝒙𝟐 + 𝟒𝒙 − 𝟓 𝟏 − 𝒙
factor the denominator
multiply the numerator and the
denominator of the 2nd fraction by -1
need the LCD for each
add
Ex. Add:
𝒙𝟐
𝒙 + 𝟑
𝒙 − 𝟏
+
+ 𝟐𝒙 − 𝟖 𝟐 − 𝒙
Ex. Add:
𝒙𝟐
−𝟑
𝒙−𝟐
−
− 𝒙 − 𝟐𝟎 𝟓 − 𝒙
Lesson 76: Solving Three Equations and Three Variables Using Substitution and
Elimination
We need the (x, y, z) that makes all three equations true.
First use substitution. Substitute for the same variable in two different
equations. This makes two equations with the same two variables. Solve them
using substitution or elimination. Find the last variable’s value.
Ex. Solve:
2x + 3y – z = 0
x+y+z=2
x + 2y = 0
Ex. Solve:
x – y + z = -7
2x + 3y – z = 0
y = 3z
Lesson 77: Advanced Equations Containing Radicals
Ex. Solve:
𝟑
�𝒙𝟑 + 𝟗𝒙𝟐 + 𝟒𝒙 + 𝟒 − 𝟑 = 𝒙
isolate the root
cube both sides
solve
Ex. Solve:
√𝒙 + 𝟏𝟐 + √𝒙 = 𝟔
need one root per side
square both sides, use FOIL on the
right
isolate remaining root
square both sides
a check is required when you square
both sides of an equation
Ex. Solve:
√𝒀 − 𝟗 − 𝟑 = √𝒀
Lesson 79: Metric Volume
Recall: units of length can be used to measure areas when squared or volumes
when cubed:
1 cm2
1 cm
1 cm
1 cm3
1 cm
3
1 cm = 1 ml
1000 ml = 1 liter
1000 cm3 = 1 liter
Ex. Use unit multipliers to convert 9000 ml to cubic inches.
Ex. Use unit multipliers to convert 8 cubic feet to liters
1 cm
1 cm
Lesson 80: Using Proportions to Solve Variation Problems
Recall:
A varies directly as B means A = kB or
A varies inversely as B means A = k
𝟏
𝑩
or
𝐴1
=
𝐵1
𝐴1
=
𝐵2
𝐴2
𝐴2
𝐵2
𝐵1
Ex. The number of worker bees in the hive varied directly as the number of
drones. If 15 worker bees were present when there were 2 drones, how many
worker bees were there when 1300 drones were present?
Ex. The illumination value of the lamp varied inversely as the square of the
distance from the lamp. If the illumination value of the lamp was 25 at a distance
of 2 feet from the lamp, what was the illumination value of the lamp at a distance
of 10 feet from the lamp?
Ex. The mouse population varied inversely as the square of the cat population.
If there were 12 mice present when there were 6 cats, how many mice were
there when 12 cats were present?
Lesson 81: Division of Complex Numbers
Do not leave i in the denominator.
Ex. Simplify:
𝟐+𝟑𝒊
𝟕−𝟐𝒊
multiply the numerator and the
denominator by the conjugate of
the denominator
i2 = -1
write as a sum of two fractions
reduce if possible
Ex. Simplify:
−𝟐𝒊 +𝟓
𝟑 −𝟒𝒊
Ex. Simplify:
𝟐 +𝟓𝒊
−𝟑𝒊 + 𝟓
Lesson 82: Advanced Complex Fractions
Ex. Simplify:
𝟏+
𝟑
𝒎
𝟐−
𝒙
𝒚
get a LCD and add 2 -
invert and multiply
get a LCD and add
invert and multiply
Ex. Simplify:
𝟑+
𝒂
𝒙
𝒌−
𝒋
𝒎
𝒙
𝒚
Ex. Simplify:
𝒙
𝒂−
𝒃
𝒄+
𝟐
𝒌
Lesson 83: Exponents Which are Variables
Recall:
x2x3 = x5
add exponents to multiply powers of x
𝒙𝟓
subtract exponents to divide powers of x
𝒙𝟐
= 𝒙𝟑
(x2)3 = x6
multiply the exponents
Ex. Simplify:
𝒂𝒙⁄𝟐 𝒃𝒚 𝒂𝟑𝒙 𝒃𝒚⁄𝟓
add exponents to multiply
fractions need the LCD to add
Ex. Simplify:
𝒙𝟐𝒂 𝒚𝒃⁄𝟑
𝒙−𝒂+𝟑 𝒚𝒃
Ex. Simplify:
(𝒙𝒎 )𝟑 𝒚𝒙+𝟐
𝒙𝟐𝒎
subtract exponents to divide
Lesson 84: Consistent, Dependent Systems
Consider:
Consistent: At least one solution
Inconsistent: No solution
Dependent: One equation is a multiple of the other, else independent.
What does the algebra look like if you solve an inconsistent (no solution)
system?
Consider: Solve y = 2x + 6
y = 2x - 3
What does the algebra look like if you solve a dependent (same line) system?
consider: Solve 2x + y = 3
4x + 2y = 6
Ex. Is the following system of equations consistent or inconsistent, dependent
or independent?
2x – 3y = 5
6y = 4x – 48
Lesson 85: One Type of Nonlinear System of Equations
Ex. Solve:
BTD – 8TD = 64
BTD + 8TD = 32
use elimination to find BTD
now find TD
now find B
Ex. Solve:
BTD + 7TD = 77
BTD - 7TD = 49
Lesson 86: Introduction to Inequalities
Consider the number line:
-6
-5
-4
-3
-2
-1
0
1
2
Numbers are larger as you go to the right.
Ex. 4 > 2 because 4 lies to the right of 2
Ex. -2 > -4 because -2 lies to the right of -4
(Remember the arrow points at the smaller value.)
Consider:
≥
means
<
≤
means
>
<
means
≥
> means ≤
Domain:
The domain consists of the allowable x values.
3
4
5
6
D = {integers} means your answer must consist only of integers, …-3, -2, -1, 0, 1,
2, 3…
D = {reals} means your answer may be any real number (any number on the
number line)
Ex. Graph x > 2, D = {integers}
-6
-5
-4
-3
-2
-1
0
1
3
2
5
4
6
Ex. Graph x > 2, D = {reals}
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
When multiplying or dividing both sides of an inequality by a negative value, flip
the inequality symbol.
Ex. Graph: -x > 7; D = {reals}
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Ex. Graph –x
-6
-5
≥ -5; D = {positive integers}
-4
-3
-2
-1
0
1
2
3
4
5
6
Lesson 87: The Slope Formula
Recall: slope = m =
𝒓𝒊𝒔𝒆
𝒓𝒖𝒏
=
𝒖𝒑
𝒅𝒐𝒘𝒏
= − 𝒓𝒊𝒈𝒉𝒕
𝒓𝒊𝒈𝒉𝒕
Ex. Find the slope of the line passing through
(-3, 4) and (2, -5).
memorize: m =
𝒚𝟐 − 𝒚𝟏
𝒙𝟐 − 𝒙𝟏
Ex. Find the slope of the line passing through
(2, -4) and (-6, 5).
Lesson 88: The Distance Formula
Recall: Find the distance between (-3, 4) and (2, -5).
memorize: d = �(𝒙𝟐 − 𝒙𝟏 )𝟐 + (𝒚𝟐 − 𝒚𝟏 )𝟐
Ex. Find the distance between (2, -4) and (-6, 5).
Lesson 89: Compound Inequalities
“And” means both conditions must be met to be included in the answer.
“Or” means include any number which satisfies either condition.
Ex. Graph
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
1
2
3
4
5
6
2
3
4
5
6
-4 > -x -2 and x + 3 ≥ 4, D = {reals}
Ex. Graph
-6
-5
-4
-3
-2
-1
0
-x - 3 < -2 or x + 3 ≤ 5, D = {integers}
Ex. Graph
-6
-5
-4
-3
-2
-1
0
1
-5 ≤ -x - 3 < -2, D = {reals}
Ex. Graph
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-x - 3 < -2 or -7 >-5 + x, D = {integers}
Lesson 90: Advanced Systems of Three Equations
Recall:
•
•
•
The solution is one ordered triple, (x, y, z), which makes all three
equations true.
First use elimination two times to eliminate the same variable from two
sets of two equations.
Then solve the resulting two equations which will have the same two
variables using either substitution or elimination.
Ex. Solve:
2x – 3y + z = 7
x + 2y – 2z = -1
-x – y + 3z = -2
Ex. Solve:
x – 2y + z = 7
2x – y – 2z = -1
3x + 5y – z = -2
Lesson 91: Systems of Linear Inequalities
Ex. Graph y < 3x + 2
the solution is all of the ordered pairs which make y < 3x +
2 true
Consider a randomly chosen ordered pair. Does it make
the inequality true?
Graph the line y = 3x + 2. It divides the plane in half. One half contains ordered
pairs which make the inequality true. Test each half to find the solution set.
Use a dotted line to graph boundary lines of inequalities with < or > to indicate
we are not including the points which lie on the line itself in the answer. Graph
boundary lines of inequalities with ≤ or ≥ with a solid line to indicate the points on
the line are included in the answer.
Ex. Graph:
y ≥ -2x + 3
y<x–2
Graph the boundary lines:
y = -2x + 3 (solid line)
y = x – 2 (dotted line)
Either: check each of the four sections of graph to see
which contains ordered pairs that make the solution true,
Or: remember y > or ≥ lies above the line while y < or ≤ lies below the line
Ex. Graph:
y≤ x+3
y>x -2
Lesson 92: Boat in the River Problems
Consider:
D = R∙T
Downstream:
Upstream:
𝒃𝒐𝒂𝒕 𝒔𝒑𝒆𝒆𝒅 𝒊𝒏 𝒄𝒖𝒓𝒓𝒆𝒏𝒕𝒔
+ 𝒔𝒑𝒆𝒆𝒅 �∙ Time Down
𝒔𝒕𝒊𝒍𝒍 𝒘𝒂𝒕𝒆𝒓
Distance Down = �
Distance Up
𝒃𝒐𝒂𝒕 𝒔𝒑𝒆𝒆𝒅 𝒊𝒏 𝒄𝒖𝒓𝒓𝒆𝒏𝒕𝒔
− 𝒔𝒑𝒆𝒆𝒅 �∙ Time Down
𝒔𝒕𝒊𝒍𝒍 𝒘𝒂𝒕𝒆𝒓
=�
DD = (B + W)TD
DU = (B - W)TU
no need to memorize these
Ex. Lydia can paddle her canoe 21 miles downstream in only 3 hours but it takes
her 5 hours to go just 15 miles upstream. What is the speed of the current? How
fast can Lydia paddle in still water?
fill in formulas
distribute
solve
Ex. Conrad’s boat travels at 30 miles per hour on a lake. He can travel 132
miles down the St. John’s River in the same amount of time he can go 108 miles
upstream. How fast does the St. John’s River flow?
Ex. The paddleboat Twain’s Folly can travel 24 miles down the Mississippi in the
same time it can travel 16 miles upstream. If the speed of the boat in still water
is 10 miles per hour, what is the speed of the current?
Lesson 93: The Discriminant
Recall:
2
The quadratic formula to solve ax + bx + c = 0, x =
−𝒃±�𝒃𝟐 −𝟒𝒂𝒄
𝟐𝒂
The discriminant = b2 – 4ac
The discriminant determines the number and types of solutions.
Ex. Solve: x2 + x + 1 = 0
If the b2 – 4ac is:
the solution(s) is (are)
negative
two nonreal complex conjugates
positive
two real numbers
zero
one real number
Ex. How many and what kinds of solutions does
3x2 – 2x = 6 have? Do not solve.
Lesson 94: An Introduction to Functions
Ex. y = 2x + 5
Ex. fare = $5 + $1∙miles
Ex. A = 50∙2t
Ex.
Math 101-1
ES 310
Math 101-2
ES 328
Math 104-1
ES 328
Math 104-2
ES 320
Def: A function is a rule of assignment that assigns to each element of the first
set (called the domain) exactly one element in the second set (called the range).
I.e. Functions are single valued. Each input (x) must be assigned to only one
output (y). Note, y values may be used repeatedly.
Ex. Is the following relationship a function?
(1, 2), (2, 3), (3, 4), (4, 5)
Ex. Is the following relationship a function?
(1, 2), (1, 3), (3, 4), (4, 5)
Ex. Is the following relationship a function?
(1, 2), (2, 2), (3, 2), (4, 2)
Ex. Is the following relationship a function?
.a
.1
.b
.2
.c
.3
.d
.4
Ex. Is the following relationship a function?
.a
.1
.b
.2
.c
.3
.d
.4
Ex. Is the following relationship a function?
.a
.1
.b
.2
.c
.3
.d
.4
Ex. Is the following relationship a function?
.a
.1
.b
.2
.c
.3
.d
.4
Function Notation:
Consider:
Ex.
y = 3x – 2
f(x) = 3x – 2
find f(3)
find g(1)
find g(-1)
Ex. h(x) = 2x2 – 1, g(x) = 3x + 5. Find h(2)
y = 5x2 + 3
g(x) = 5x2 + 3
plug in 3 for x, find y
Lesson 96: Joint and Combined Variation
Recall:
A varies directly as B means A = kB or
A varies inversely as B means A = k
𝟏
𝑩
or
𝑨𝟏
𝑨𝟐
=
𝑨𝟏
𝑨𝟐
A varies jointly as B and C means A = kBC or
𝑩𝟏
=
𝑩𝟐
𝑨𝟏
𝑨𝟐
𝑩𝟐
𝑩𝟏
=
𝑩𝟏 𝑪𝟏
𝑩𝟐 𝑪𝟐
Ex. Leonard’s Pay varied directly as the number of hours he worked and
inversely as his fatigue level. If he earned $50 when he worked 10 hours and his
fatigue level was at 2, what how many hours would he have to work to earn $150
when his fatigue level was at 1?
Ex. The level of noise at the birthday party varied directly as the number of boys
present and inversely as the number of mothers present. If 21 boys with 3
mothers resulted in a noise level of 70 decibels, how many mothers were
present when there were 10 boys and a noise level of 50 decibels?
Lesson 97: Solving Using Substitution With Fractions
Ex. Solve using substitution:
5x – 2y = 35
3x + 7y = -20
solve for x
substitute
distribute
multiply by the LCD to cancel
fractions
solve for y
find x
Ex. Solve using substitution:
2x + 3y = 5
3x – 5y = 36
Lesson 99: An Introduction to Absolute Value Inequalities
larger
Consider:
-2 > -4
The number
line:
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
The absolute
value of
numbers
on the number
...6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6..
line:
larger larger
|-4| > |-2|
The absolute value of a number is its distance from 0 on the number line.
Consider:
Ex.
|x|≤2
vs.
Ex. | x | ≥ 2
Ex. Graph:
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
-| x | + 3 < 0, D = {integers}
solve for | x |
graph solution:
for| x | < graph central
portion of number line
for | x | > graph outer
portions of number line
Ex. Graph:
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
0
1
2
3
4
5
6
-| x | + 5 ≥ 1, D = {reals}
Ex. Graph:
-6 -5 -4 -3 -2 -1
| x | - 4 > 0, D = {integers}
Lesson 99B: Factoring by Grouping
For factoring polynomials containing 4 terms:
Ex. Factor:
ax + ay + 3x + 3y
group as 2 sets of 2 terms
factor the GCF off each set
if binomials match, then continue
factor matching binomial off
check if desired
Ex. Factor:
-9x + 9a + 3x – 3a
Ex. Factor:
4a – 4b – ca + cb
Lesson 100: Graphing Parabolas
Ex. y = -(x - 3)2 + 2
opposite x
same y
vertex (3, 2)
x y
Notice: y = -(x – 3)2 + 2 is the same as y = -x2 + 6x – 7
Parabolas of the form y = ax2 + bx + c open:
•
•
downwards if a is negative
upwards if a is positive
Ex. Complete the square as an aid to graphing:
y = x2 – 4x + 1
move constant to the other side
make binomials (x – k)(x – k) where k
is ½ the x coefficient
add k2 to left
isolate y
identify vertex
find other points using a T chart
graph
Ex. Complete the square as an aid to graphing:
y = -x2 + 6x - 4
Ex. Complete the square as an aid to graphing:
y = -x2 - 2x + 1
Lesson 101: Percent Markups
Selling Price = Purchase Price + Markup
Sp = PP + M
Recall:
Write 78% as 0.78,
5.5% as 0.055
Move the decimal 2 places to the left
Ex. The selling price of the pickup was $28,050.00. If the markup was 10% of
the purchase price, how much did the dealership pay to purchase the pickup?
Ex. The Jewelry representative was selling the tiger’s eye necklace for $156.00.
If she had marked it up 30% of her purchase price, what was her purchase
price?
Lesson 102: Sums and Products of Functions
Ex. f(x) = x + 3, D = {reals}.
g(x) = x2 + 2, D = {integers}.
a)
b)
c)
Find (f + g)(2).
Find (f + g)(3).
Find (f + g)(x).
Find f(2), g(2)
Add the results
Ex. h(x) = 2x – 1, D = {positive integers}.
g(x) = x2 + 3, D = {reals}.
a)
b)
c)
Find hg(3)
Find hg(-2)
Find hg(x)
Ex. f(x) = 3x – 5, D = {reals}.
h(x) = x2 – 6, D = {positive integers}
a)
b)
c)
Find (f + h)(3)
Find (f + h)(-2)
Find (f + h)(x)
Lesson 105: Advanced Factoring
Consider: (2x + 5)(3x – 2)
Ex. Factor -11x + 6x2 – 10
Rearrange in descending order: ax2 + bx + c
Find the key number a∙c
Find two numbers that multiply to equal the key
number and whose sum is the x coefficient, b.
Rewrite the middle term as a sum of two terms
Factor by grouping
check
Ex. Solve
-33x – 7 = -10x2
Ex. Solve: -14x2 – 6x + 12x3 = 0
Lesson 106: More on Systems of 3 Equations
Remember, the initial goal is to use elimination to attain 2 equations with the
same 2 variables. In other words, choose a variable to eliminate from 2
equations. The resulting equation and the equation you did not use will be two
equations with the same two variables. Use substitution or elimination to solve
these two equations.
Ex. Solve:
2x – 5y = 7
x + 2z = 5
3y – z = -5
Ex. Solve:
x–y=1
2x + z = 4
y – 2z = 1
Lesson 108: Factoring the Sum and Difference of Cubes
Memorize:
a3 + b3 = (a + b)(a2 - ab + b2)
a3 - b3 = (a - b)(a2 + ab + b2)
check:
Notice: a3 + b3 ≠ (a + b)(a + b)(a + b)
(a2 + 2ab + b2)(a + b)
a3 + 2a2b + ab2 + a2b + 2ab2 + b3
a3 + 3a2b + 3ab2 + b3
Ex. Factor
27x3b6 - 8y9
what is being cubed
(
-
)(
+
+
)
Ex. Factor
64z6m9 + 125p3k6
Ex. Factor
8y6 - 27x15j3
Lesson 109: More on Fractional Exponents
Ex. Simplify:
(2y1/2z1/3x)3
Ex. Simplify
(x1/4 + z1/2)2
multiply the binomial times itself
use FOIL to distribute
multiply powers of x by adding exponents
add like terms
Ex. Simplify
(x1/2 + y1/3)2
Lesson 110, 112: Quadratic Inequalities
Ex. Solve:
(x – 5)(x + 4) > 0; D = {reals}
set each factor = 0
solve
label number line sections
-6 -5 -4 -3 -2 -1 0
test
section
1
2
3
4
5
6
?
number (x – 5)(x + 4) > 0
check each section of number
line to see if it contains solutions
Ex. Solve:
x2 + 3x ≤ 10; D = {integers}
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Ex. Solve:
x2 + 6 ≤ -5x
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Lesson 111: Word Problems With Three Equations
Ex. There were 38 quarters, dimes and nickels in all, and their value was $6.35.
How many coins of each type were there if there were 6 times as many quarters
as nickels?
Ex. There were 31 soft drinks in all. A large contained 30 ounces, a medium
contained 12 ounces and a small contained 10 ounces. In fact, altogether the
soft drinks contained a total of 560 ounces. If there were 4 more medium drinks
than large drinks, how many of each size were there?
Ex. Jamison’s woefully small piggybank contained just $4.65 in quarters, dimes
and nickels. There were 51 coins altogether including 6 times as many nickels
as dimes. How many of each kind of coin was there?
Lesson 114: (substituted) Absolute Value Equations
Ex. Solve |x| + 3 = 10
isolate absolute value
find the two solutions
Ex. Solve:
|x – 2| + 3 = 12
Ex. Solve:
|2x – 1| + 5 = 16
Lesson 117: Set Notation
•
•
•
•
ℝ = {reals numbers}, any number which can be placed on the number line
ℂ = {complex numbers}, any number which can be written in the form a + bi
where a and b are real numbers
ℤ = {integers}, …-3, -2, -1, 0, 1, 2, 3,…
ℝ2 = {(x, y)| x, y ∈ ℝ}, the ordered pairs in the Cartesian Plane
Ex. Graph A = {x ∈ ℤ | -2x + 5 < 10}
solve the inequality
graph the integers which
are solutions
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Ex. Graph
B = { x ∈ ℝ | x2 ≤ 4x + 12}
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Ex. Graph
C = { (x, y) ∈ ℝ2 | y ≥ 2x – 3 and x > -2}
Lesson 119: Absolute Value Inequalities Part 2
Recall:
larger
Consider:
-2 > -4
The number
line:
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
The absolute
value of
numbers
on the number
...6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6 ..
line:
larger
larger
|-4| > |-2|
The absolute value of a number is its distance from 0 on the number line.
Consider:
Ex.
|x|≤2
vs.
Ex. | x | ≥ 2
Ex. Graph: {x ∈ ℤ | |x – 3| ≤ 5}
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Ex. Graph: {x ∈ ℝ | |x + 2| ≥ 1}
-6 -5 -4 -3 -2 -1
0
1
2
3
4
0
1
2
3
4
5
6
Ex. Graph: {x ∈ ℤ | |x - 1| ≥ 4}
-6 -5 -4 -3 -2 -1
5
6
Lesson 121: Rational Inequalities
Ex. Graph: 2 ≤
3
𝑥+4
need 0 on one side
need a single fraction
on the other side, get
LCD and combine
zeros of the numerator
and the denominator
divide the number line
into sections
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
section
test
check each section of
number
number line to see if it
contains solutions
Ex. Graph:
1
>
𝑥+2
𝑥−3
-6 -5 -4 -3 -2 -1
0
1
3
2
5
4
6
Ex. Graph:
2
≥
𝑥
𝑥+1
-6 -5 -4 -3 -2 -1
0
1
2
3
4
5
6
Lesson 131: More on Function Notation
Recall:
If f(x) = 3x - 1, D = {reals}, then f(2) = 5.
f(10) = 29
i.e. we replace x with the value in the parenthesis
Ex. If f(x) = 3x - 1, D = {reals}
a. find f(7)
b. find f(a)
c. find f(2a)
d. find f(x + 1)
Ex. If g(x) = x2 - x, D = {reals}
a. find f(7)
b. find f(a)
c. find f(2a)
d. find f(x + 1)
Ex. If h(x) = 2x - x2, D = {reals}
a. find f(7)
b. find f(a)
c. find f(2a)
d. find f(x + 1)
Lesson 132: An Introduction to the Graph of the Absolute Value Function
Ex. Graph y = | x |
xy
Many graphs of absolute value functions are V shaped. All of the graphs we will
be graphing in the class are. Make sure you choose enough x values to find the
V shape.
Ex. Graph y = | x | - 3
xy
Ex. Graph y = | 2x - 1 |
xy