Lesson 1.01
Part I: Evaluating
Exponents represent how many time to multiply a number starting with 1.
e.g.
2! = 1 ∙ 2 ∙ 2 ∙ 2 ∙ 2
= 16
You need to memorize all of your squares and double. A square is a number
multiplied by itself. A double is a number added to itself.
Part II: Equations
To solve an equation we must use inverse operations. In order to know the inverse
operations, you must know the most important thing in all of Algebra:
The Orders of Operation:
1.
+/−
2. ×/÷
3. Exp./Roots
This will be the key to every important concept in Algebra.
Operations written on the same line are inverses, meaning they undo or reverse one
another. So, to reverse addition we use subtraction and vice-versa. To reverse
multiplication we use division. To reverse exponents, we use roots.
We must get the ! by itself by getting rid of all the other numbers on the same side
as the !. So, for instance in the equation:
!+5=9
We must get rid of the 5, so that ! will be by itself. In order to get rid of a number,
we must first figure out what it’s doing. The 5 is adding. So, to get rid of it we must
subtract 5. One of the most important things to remember is:
!ℎ!"#$#% !"# !" !" !"# !"#$ !" !"#$%&'(, !"# !"#$ !" !" !ℎ! !"ℎ!".
So, if we subtract 5 from the left side, we also subtract 5 from the right side:
!+5 = 9
−5 − 5
!
= 4
If we plug 4 back into the original equation, we will see that it works:
4+5=9
Lesson 1.01
Part III: Functions
Functions are made up of ordered pairs. An ordered pair is different than just a
single number, which you have been used to working with up until now. An order
pair is a pair of numbers, in which the order matters. We always write ordered
pairs using parentheses. So, if we want to talk about the pair where 2 goes first and
then 3, we write (2,3), which is different than (3,2). We call the first number !
and the second number !.
A function is like a rule that tells us how to go from one number to another. So for
instance, a function rule might tell you always to add 1 to your first number to get
the second number. It might say to double the first number to get the second
number. The possibilities are endless and analyzing these types of rules will be a
fundamental part to your further studies of math.
This function says to add 2 to the first number (!) to get the second number (!).
! =!+2
Then, we can plug in any number for the ! to figure out the !. This can be written as
a table. We can then graph the ordered pairs we get. The first number, x, tells us how
much to move to the left/right. The second number, y, tells us how much to move
up/down. Positive number go to the right or up. Negative numbers go to the left or
down.
!
!
Ordered pair
0
2
(0,2)
1
3
(1,3)
2
4
(2,4)
3
5
(3,5)
4
6
(4,6)
5
7
(5,7)
6
8
(6,8)
Lesson 1.02
Part I: Evaluating
Adding a number to itself is the same as multiplying it by 2.
e.g.
5+5=2∙5
This works with any number. We can use ! to represent that it works for any
number, even if we don’t know what it is:
e.g.
! + ! = 2!
Likewise, multiplying a number by itself is the same as raising it to the exponent of 2.
e.g.
5 ∙ 5 = 5!
This works with any number. So, just as before we can write:
e.g.
! ∙ ! = !!
Part II: Equations
In the last lesson we only solved equations with order 1 operations, i.e. addition and
subtraction. Now, we will do the same thing but with multiplication or division. Just
as addition cancels out subtraction, multiplication cancels out division and viceversa.
e.g.
3! = 15
We want to get x by itself. So, we must get rid of the 3. In order to get rid of a number,
we must find out what it’s doing. The 3 here is multiplying. So, we must divide by 3 to
get rid of the 3, which is multiplying.
!!
The 3’s cross out, and we get:
!
!"
= !
!=5
Part III: Functions
The only thing different with the functions for this lesson is to be able to multiply by
a number instead of adding a number. But, you do the exact same thing as in the
previous homework. You find the ordered pairs by multiplying and then graph them.
Lesson 1.03
Part I: Evaluating - Addition and Subtraction
Adding two positive numbers is very easy. You start at the first number an then go
up by whatever the second number is.
e.g.
5+3=8
Adding a positive number and a negative number is also rather easy. When you add
a negative number to a positive number, instead of going up you go down.
e.g.
5 + −3 = 2
Subtraction is the inverse operation of addition. It finds what number to add to the
second number to get to the first number.
e.g.
7−3
To figure this out we need to figure out what number you add to 3 to get back to 7.
Since 3 + 4 = 7, we know 7 − 3 = 4.
Sometimes in order to get from the second number to the first you need to go down.
So, the answer will be a negative.
e.g.
3 − 7 = −4
This works because in order to get from 7 to 3 you need to go down 4.
Combining
When you combine when adding or subtracting the multiple will changes.
When you combine when multiplying or dividing the exponent changes.
When there is no multiple or exponent with an !, you can think of it as 1! or ! !
e.g.
3! + !
3! + 1!
4!
Lesson 1.04
Part I: Evaluating - The Order of Evaluation
When more than one operation is written in an expression, we need to know which
one do to first.
e.g.
3+5∙2
Do we do 5 + 3 first or do we do 5 ∙ 2 first?
It’s a very important question, because we’ll get different values depending on
which one we do first.
As with most things in algebra, the way to figure it out is based on:
The Orders of Operation:
1.
+/−
2. ×/÷
3. Exp./Roots
The rule is that you must always perform the bigger operations first (unless
there is a grouping symbol). This means that you will perform all the exponents and
roots first, and then all the multiplication and division, and then all the addition and
subtraction.
Following this rule means that for the above example you must do:
3+5∙2
3 + 10
13
If two operations are the same size, we do them in the order they appear from
left to right.
e.g.
7−6+2
1 + 2
3
If there is a grouping symbol, then we figure out the value inside the grouping
symbol first. What is a grouping symbol? Parentheses, fraction bars, exponents, and
root symbols are all examples of grouping symbols.
e.g.
3(5 + 2)
We must do 5 + 2 first because it is in a grouping symbol.
e.g.
!!!
!
We must do the 5 + 1 first because it is being grouped by the fraction bar.
e.g.
2!!!
We must do 3 + 1 first because it is being grouped as an exponent.
Lesson 1.05
Part I: Evaluating - Multiplying by a fraction
When you multiply by a unit fraction, you simply divide by the denominator.
e.g.
!
12 ∙ !
!"
!
3
Part II: Solving Equations - Socks and Shoes
So far, we’ve only had to solve equations with one number on the same side as the !.
These types only take one step to solve. Now, we will do equation with two numbers
on the same side as the !. These will take two steps to solve. The important thing to
figure out is: Which number do you get rid of first? The answer is:
You take off first, whatever is put on last.
This is the socks and shoes principle. You have to take your shoes off first, because
you put them on last.
So, in order to know which number to get rid of first, we need to know which one
happens last. In order to know that, we need to know the Order of Evaluation.
(Remember, bigger operations go first, and we need to do whatever is in a grouping
symbol first.) Then, once we know what happens last, we simply get rid of it by
using the inverse operation.
e.g.
3! + 5 = 14
In this case, we need to get rid of two numbers, the 3 and the 5. So, we need to ask,
which one is happening last? Well, if we imagine that ! is any other number, which
operation would be performed last? First, we would multiply by 3 and then add 5,
which means we need to get rid of the 5 first and then the 3. How do we get rid of
the 5? We need to first find out what it is doing. The 5 is adding, so to get rid of it we
need to use the inverse operation, subtraction. Therefore, we must subtract 5 from
both sides.
3! + 5 = 14
−5 − 5
3!
=9
Then, we get rid of the 3.
3! = 9
3 3
!=3
Now we’re done because the ! is written by itself. We can plug ! = 3 back into the
original equation to check that it works:
3 ∙ 3 + 5 = 14
Lesson 1.06
Part I: Evaluating - Negative of a Negative
The negative of a negative is a positive.
e.g.
− −4 = 4
So, if you have three negatives, two of them will become a positive, and you’ll just be
left with a negative.
e.g.
− − −4 = −4
Part II: Solving Equations - Getting rid of a negative
When solving equations, sometimes we need to get rid of the negative sign, rather
than a negative number. In order to cancel out the negative sign, we need to just
multiply by a negative sign again, since a negative of a negative is a positive.
e.g.
−! = 4
In order to get ! by itself, we need to get rid of the negative sign. So, we multiply by
a negative on both sides.
− −! = − 4
The negative times a negative cancels to become a positive, and we get:
! = −4
Negative sign is multiplication
A negative sign is performing the operation of multiplication. So, when doing the
orders of evaluation, we treat the negative sign as an order 2 operation.
e.g.
−! + 2 = 5
The negative sign happens first, then the +2. So, we get rid of the 2 first.
−! + 2 = 5
−2 = −2
−!
= 3
Then, we get rid of the negative by multiplying by a negative on both sides:
− −! = − 3
! = −3
Lesson 1.07
Part I: Simplify and Combine
This lesson introduces one of the most important concepts in Algebra:
The Double-Switch Rule for Addition and Subtraction.
When you subtract a negative, you add a positive.
When you subtract a positive, you add a negative.
Later, you will learn that there is a double-switch rule for Multiplication-Division
and one for Exponents-Roots. (In fact, you already did the Multiplication-Division
one in the Lesson 1.05 - Multiplying by a unit fraction.) But for now, we’ll just use
the one for Addition-Subtraction.
Although it might not be clear as to why we would want to switch out subtraction
for addition, there are many advantages in Algebra in doing so. Oftentimes,
switching out subtraction for addition will enable us to figure problems out that we
otherwise would not be able to.
Here’s an example in evaluating:
e.g.
5 − (−2)
We can figure this out by doing normal subtraction and finding what we have to add
to −2 to get to 5, which is obviously 7, or we could use the double-switch rule to
change it into addition:
5 − −2
5 + +2
7
Notice that you get the same answer whether you figure it out by subtraction, or
whether you figure it out by using the double-switch rule. Many people find it more
useful to just always do addition rather than to use subtraction, but it is very good to
know how to do it both ways.
Here’s another example:
−7 − 2
We can either think of this as subtraction, i.e. what do I add to positive 2 to get
negative 7? And we get −9. Or, I can think of it as:
−7 + −2
This is saying start at −7 and go down 2. So, we get −9. Notice that either way, you
end up with the same answer.
Lesson 1.07
Part II: Solving Equations
Sometimes we need to use the double-switch rule to solve equations. An example is:
7−! =5
Although it should be rather obvious that the answer is ! = 2 since 7 − 2 = 5, it is
very important to know how to solve this algebraically, since there will be many
problems that we will have that are not so easy as this one.
The problem is that if we try to follow the normal procedure, we would need to get
rid of the 7. But, in order to get rid of a number, we need to know what it is doing.
The 7 isn’t doing anything. Be very careful here. Many students at first are tricked
into adding 7 to both sides. But,
The 7 is not subtracting, but rather being subtracted from.
Make sure that you understand this before moving on.
If we double-switch though, we can change the subtraction sign ino an addition sign.
7 + −! = 5
Now, because addition is commutative, i.e. we can switch the order around without
changing the value, we get:
−! + 7 = 5
Notice that in this equation the 7 is now doing something. It is adding. So, we simply
subtract it to get the ! by itself:
−! + 7 = 5
−7 = −7
−!
= −2
Then, just multiply by a negative:
− −! = − −2
!=2
This is a very important technique. Whenever you have to solve an equation where
there’s an something minus a term with ! in it, you need to double-switch and then
solve.
Part III: Functions
Whenever you plug in an !, you always want to plug it in in parentheses.
e.g.
! = −! + 3
To plug in ! = −4, we put it in parentheses.:
! = − −4 + 3
!= 4 + 3
!= 7
So, we get the ordered pair (−4, 7)
L 1.08
The homework for this lesson is all review.
The skills practice is on converting between mixed numbers and common fractions.
A common fraction is a fraction that is reduced whether or not the numerator is
larger than the denominator. In algebra, we usually want to use common fractions
instead of mixed numbers. From now on, only rarely will we use mixed numbers.
To convert from mixed number to common fraction simply multiply the whole
number part by the denominator, add that number to the numerator, and then write
that new number over the denominator.
e.g.
!
3!
!"!!
!
!"
!
To go the other way, just divide the top by the bottom and then write the remainder
over the denominator. This is actually a form of distribution, which we’ll learn about
more later:
!"
!
!"!!
!
!
4!
Lesson 1.09
Part I: Evaluating - Multiplying and dividing by negatives
Multiplying by negatives is very easy once you realize it’s all based upon the idea
that a negative times a negative is a positive. Whenever you multiply a ne..g.egative
times a negative, they cancel eachother out and you get a positive:
e.g.
−5 ∙ −4
Because you are multiplying two negatives, they simply cancel out and you get:
20
If there’s only one negative in multiplication, you’ll just end up with a negative:
e.g.
−5 ∙ 4 = −20
e.g.
5 ∙ −4 = −20
Once you know how to multiply negatives, dividing is easy. Since dividing is just the
reverse of multiplication, the same rule applies. If you divide a negative by a
negative, you get a positive. If you only have one negative in the division, then you
will just end up with the answer being negative.
e.g.
!!"
!!
Since it is a negative divided by a negative, they cancel out and you get:
5
This makes sense because in every division problem you need to find what number
we multiply the bottom by to get the top number. −4 ∙ 5 = −20, so it works.
e.g.
!!"
!
Because there is only one negative in the division, you will end up with a negative:
−4
This makes sense because 5 ∙ −4 = −20.
e.g.
20
−4
Again, there is only one negative, so you will end up with a negative:
−5
This makes sense because −4 ∙ −5 = 20.
Lesson 1.10
The homework is all review.
Multiplying one digit by two digit numbers:
When mulitplying a two digit number by a one digit number mentally, you always
work from left to right.
e.g.
11×6 = 10×6 + 1×6
= 60 + 6
= 66
e.g.
13×5 = 10×5 + 3×5
= 50 + 15
= 65
You should bet able to quickly multiply any number between 1 and 30 by any
number bettwen 1 and 10.
Lesson 1.11
Part I: Evaluating- Roots
Roots are the invereses of exponents. The index of the root reverses the exponent of
the base.
e.g.
!
8
The index is 3. So, to figure this out we need to figure out what base with an
exponent of 3 would be equal to 8. The answer is 2 becae 2! = 8
e.g.
!
25 = 5
Becase 5! = 25
If there is no number written in the index, then that means the index is 2.
Part II: Equation - Roots and Exponents
Just like we use addition to get rid of subtraction and multiplication to get rid of
division, we will use roots to get rid of exponents and vice versa.
e.g.
!! = 8
To get rid of the exponent we take the root of each side with 3.
!
!
!! = 8
!=2
e.g.
!
!=5
To get rid of the root, we take the exponent of each side with 2.
!
!
! = 5!
! = 25
Lesson 1.12
Part I: Evaluating - Negative Bases
This lesson introduces negative bases. A base is the thing the is being raised to an
exponent. Like in:
5!
5 is the base.
3 is the exponent.
In order to have a negative base, you must use parenthese around the base.
−4 !
−4 is the base.
2 is the exponent
Many students find this tricky. If you don’t write the parentheses, then the negative
sign is not part of the base:
−4!
4 is the base, not −4
2 is the power
The negative comes in after you square the base, 4.
So,
−4 ! = −4 ∙ −4 = 16
whereas
−4! = −4 ∙ 4 = −16
Part II: Solving Equations
Notice that whether you square 4 or −4 you get the same answer, 16. Because of
this, whenever you solve an equation with a square, you must include both of these
possible solutions in your anwer:
e.g.
! ! = 16
Up until now, you’re only written 4, but for now on you must write ±4, which is read
“plus or minus four”, and is a short hand for writing “4 or −4”. You must write this
because both 4 and −4 work as solutions. Whenever you solve an equation, you
must write all possible solutions.
Whenever you have an even exponent, any base will give the same value whether
it’s positive or negative. So, here’s a very important rule you must remember:
!ℎ!"!#!$ !"# !"# !"# !" !" !"!# !"#$%!"# !ℎ!" !"#$%&' !" !"#$%&'(,
!"# !"#$ !"#$% ±
This will take you a bit to get used to. Many students tend to forget this for a while
before it becomes a habit. Do your best to remember and to understand it.
Part III: Functions
Whenever you plug a number in for a variable, you must put it in parentheses.
e.g.
! = !!
Evalute for ! = −4
When you plug it it, it must be in parentheses:
! = −4 !
! = 16
Lesson 1.13
Part I: Simplify and Combine - Combining through Multiplication
So far we have done combining through addition and subtraction:
e.g.
2! + 3! = 5!
e.g.
2! + ! = 3!
Now, we will do combining through multiplication.
The key thing to remember is:
! + ! = 2!
! ∙ ! = !!
If you remember those two things, it will clear up a lot of confusion.
Because multiplication counts repeated addition or subtraction, the multiple will
change when you add or subtract. But exponents count repeated multiplication. So,
the exponent will change when you are multiplying.
Notice the similarities in the following two equations:
2! + 3! = 5!
and
!! ∙ !! = !!
Here’s another good spot to look for similarities:
3! + ! = 4!
and
!! ∙ ! = !!
Part II: Solving Equations - Get the variable written once
There are only two steps to solve any equation.
1. Get the variable written once.
2. Get the variable by itself.
So far, all of the equation we have solved have had step 1 completed, i.e. the variable
was already written once. In this lesson’s homework, therere are problems where
the variable is not yet written once, like:
3! + ! = 12
Notice, that we can’t possibly go onto step 2, getting the variable by itself, until we
have the variable written once. In fact step 1 is by far the hardest part of solving
equations. Once the variable is written once, it is always very easy to solve the
equation. There are equations where the variable is not written once that are actual
completely impossible to solve.
So, how do we get the variable written once? For now and for a while the way to get
the variale written once is through combining, which you already know how to do.
So, for instance in the above example we will first combine:
3! + ! = 12
4! = 12
After the variable is written once, now we can solve as usual, by using the socks and
shoes principle:
4! = 12
4
4
!=3
Lesson 1.13
Part III: Functions - Combining first
Same thing with graphing functions. It is always easier to graph the function is you
get the variable written once first through combining.
e.g.
If you have to graph:
! = 2! + !
combine first to make it:
! = 3!
Then plug numbers in and graph it.
Lesson 1.14
Part I: Simplifying and Combining - +/- fractions with the same denominators
Whenever you combine two terms in algebra by adding or subtracting, they must be
“like terms”. This means that everything is the same except the multiple.
e.g.
3! and 2! are like terms because the only thing different is the multiple
5! ! and 3! ! are like terms
2! ! and 3! are not like terms, because they have diferent exponents.
It’s no different with fractions. With a fraction the multiple is written in the
numerator:
!
!
=
3
!
!
So, in order to combine two fraction, we need to make sure that the only thing that is
different is the top part of the fraction, i.e. the numerator. Another way of saying this
is that you must have the same bottom number, i.e. denominator.
So, take for instance:
!
!
+!
!
Because the bottom part of each fraction is the same, we can combine them by
adding the multiples, i.e. the top part of each.
!
!
!
+!=!
!
Reducing a fraction:
Reducing a fraction is when we simplify a fraction by finding a number that divides
evenly into both the top and the bottom and then cancel it out.
e.g.
!"
!"
5 goes into both 10 and 15. So, we can rewrite it as:
!∙!
Then, we cancel the 5’s.
!∙!
!
!
Sometimes after you add or subtract two fractions, it becomes possible to reduce the
fraction. You must always reduce your fractions in your final answer.
e.g.
!
!
− !"
!"
!
!"
!
!
L 1.15 Distribution
How do you evaluate 8 ∙ 11?
Most people do 8 ∙ 10 + 8. This is what in Algebra we call distribution.
It works because 8 ∙ 11 = 8 10 + 1
= 8 ∙ 10 + 8 ∙ 1
= 80 + 8
= 88
In Algebra, you will have many problems like:
3(! + 5)
In order to simplify them, we will distibute by applying the operation on the outside
of the group to each number inside the group. So, here, we get:
3! + 15
Another example is:
5 ! − 7 = 5! − 35
We can also distribute exponents:
3 ∙ 4 ! = 9 ∙ 16
12! = 144
Notice we get the same answer whether we evaluate the inside first, like usual, or
whether we distibute the exponent to each term on the inside.
But, distribution does not always work. Consider:
2+5 !
If we tried to distribute, we would get:
4 + 25 = 39
But, if we evaluate what’s in the grouping symbol first, we get:
7! = 49
So, distribution does not always work.
L 1.15 Distribution
In order to properly distribute you must remember the most fundamental concept
in all of Algebra, the Orders of Operation.
1.
+/−
2. ×/÷
3. Exp./Roots
Many people get confused as to when distribution is allowed because they forget the
Orders of Operations. If you remember the Orders of Operations, then you will
always be able to remember whether distribution is allowed or not.
The Distribution Rule:
Distribution is only allowed when going from 3 to 2 or from 2 to 1.
So, in the first few examples, distribution worked because we were following the
rule.
In the first example:
3 ! + 5 = 3! + 15
we are going from multiplication (order 2) to addition(order 1). So, it works.
In the second example:
5 ! − 7 = 5! − 35
we are going from multiplication (order 2) to subtraction (order 1). So, again, it
works.
In the third example:
3 ∙ 4 ! = 9 ∙ 16
We are going from exponents (order 3) to multiplication (order 2). So, it works.
In the last example:
2 + 5 ! ≠ 4 + 25
We are trying to go from exponents (order 3) to addition (order 1). So, it doesn’t
work.
What about 8 10 + 1 ? This is multiplication to addition. So, the trick most people
use to multiply does work.
!
What about 64 ∙ 121 ? If we try to do it the old way, we would have to multiply 64
by 121 and then take the square root, but with distribution it is easy.
You just do 8 ∙ 11 = 88. Now, try doing worksheet 1.15 to practice distributing.
Lesson 1.16
Part I: Simplify - Double-switch then distribute
In the last lesson, you learned how to distribute. When there is subtraction, though,
you must first double-switch before you distribute.
e.g.
−3 ! − 6
First, double-switch:
−3 ! + −6
Then, distribute:
−3! + 18
Here’s another example:
− !−5
Double-switch:
− ! + −5
Then, distribute:
−! + 5
Part II: Solving equations - Fractions
When you need to get rid of a fraction that is multiplying, you multiply by its
reciprocal.
e.g.
!
!=6
!
!
!
!
The reciprocal of ! is ! . So, we multiply by ! on both sides:
! !
!
∙ ! =6∙!
! !
!=9
Lesson 1.17
Part I: Simplifying and Combining - +/- fractions with different denominators
Whenever you combine two terms in algebra by adding or subtracting, they must be
“like terms”. This means that everything is the same except the multiple.
e.g.
3! and 2! are like terms because the only thing different is the multiple
5! ! and 3! ! are like terms
2! ! and 3! are not like terms, because they have diferent exponents.
It’s no different with fractions. With a fraction the multiple is written in the
numerator:
!
!
=
3
!
!
So, in order to combine two fraction, we need to make sure that the only thing that is
different is the top part of the fraction, i.e. the numerator. Another way of saying this
is that you must have the same bottom number, i.e. denominator.
So, take for instance:
!
!
+
!
!
The 3 and 5 in the bottom of the fractions are not the same. So, in order to combine
the two fractions, we must first make them equal. We do this by first finding a
number that both 3 and 5 can go into, i.e. the least common multiple. Then we
multiply the bottom of each fraction to equal that number and also multiply the top
by the same number:
!∙!
!∙!
+ !∙!
!∙!
!"
!"
+ !"
!"
!!
!"
We do the same thing when we need to subtract two fractions:
e.g.
!
!
−
!
!
The number that both 4 and 2 go into is 4. So, we change the bottom of each fraction
to 4 by multiplying. The first one is already 4, so we leave that alone.
!∙!
!∙!
!"
!
!
−!
!
−!
!
!
Lesson 1.17
Part II: Solving Equations - Variables on both sides
Remember that the first step to solving an equation is to get the variable written
once. So, if the variable is on both sides of the equation, then obviously it is not yet
written once. In order to get it written once, you must get rid of the variables on one
side.
e.g.
5! + 2 = ! + 10
To solve this, we will get rid of the ! on the right side of the equation by subtracting
! from both sides. We could also subtract 3! from both sides. Either way will work.
5! + 2 = ! + 10
−!
−!
4! + 2 = 10
Now, the ! is written once, so we finish solving by getting ! by itself by using the
socks and shoes principle.
4! + 2 = 10
−2 − 2
4! = 8
4
4
! = 2
Here’s another example:
5! + 7 = −2! + 13
We need to get rid of the !’s on one side. So we will add 2! to both sides. We could
also subtract 5! from both sides. Either way will work.
5! − 1 = −2! + 13
+2!
+ 2!
7! − 1 =
13
Now that ! is written once, we solve by using the socks and shoes principle:
7! − 1 = 13
+1 + 1
7!
7
=
14
7
!=2
Part III: Functions - Graphing ! = ! !
The only tricky thing about this is making sure that when you plug in a number for
!, you always put the number in parentheses.
e.g.
! = !!
Plug in ! = −3
! = −3 !
!=9
So, we have (−3, 9) as our ordered pair. If you don’t put it in parentheses, then you’ll
get ! = −9, which is incorrect.
Lesson 1.17
Lesson 1.18
Part I: Simplifying
A monomial is an expression with a variable that has a multiple and an exponent.
e.g.
4! !
There doesn’t need to be an exponent written.
e.g.
3! is also a monomial, we can think of the exponent as 1, i.e. 3! !
Oftentimes, in Algebra, we will be adding or subtracting monomials. If there is the
same variable with the same exponent, they are called like terms, even if the
multiple is different.
e.g.
4! ! and 7! ! are like terms
If we add or subtract monomials that are like terms, then we can combine them into
one monomial by simply adding or subtracting the multiple.
e.g.
4! ! + 7! !
11! !
Notice that the exponent never changes when you add or subtract. This is because
exponents keep track of repeated multiplication, not repeated addition.
We can also do this when subtracting:
e.g.
4! ! − 7! !
−3! !
As long as you have the same variables with the same exponent, you can combine
the two monomials like this. If they are different variables or different exponents,
then you cannot, and must leave the expression as it is.
We can also combine monomials through multiplication. When you combine
through multiplication, it doesn’t matter if they are like terms. As long as you have
the same variable, you can combine the two monomials.
e.g.
3! ! ∙ 4! !
We can rearrange and combine:
3 ∙ 4 ∙ !! ∙ !!
Here’s another example:
Rearrange and then combine:
12! !
4! ∙ 2! !
4 ∙ 2 ∙ ! ∙ !!
8! !
Normally we just rearrange it in our heads without writing that step, but if you’re
getting stuck, just write it out.
Lesson 1.19
Part I: Simplifying - Exponent of an expression with an exponent
In this lesson we will figure out how to apply an exponent to an expression that
already has an exponent.
e.g.
!! !
Notice that this is the same as:
!! ∙ !!
We learned in lesson 1.13 that this is:
!!
Another example:
!! !
This is the same as:
!! ∙ !! ∙ !!
Which is equal to:
! !"
In general if we have an expression with an exponent raised to an exponent, we can
just multiply the two exponents.
e.g.
!! !
! !"
Part III: Functions - Slope and y-intercept
The slope is how quickly the graph goes up, i.e. how much the graph goes up every
time the ! value goes to the right one unit. This will always be whatever the multiple
of the ! is.
e.g.
! = 3! + 1
The slope is 3 because that’s the multiple of the !.
e.g.
! =!−4
The slope is 1 because that’s the multiple of the !.
The y-intercept is where the graph crosses the y-axis. This happens whenever ! = 0.
So, to find the y-intecerpt you just plug in ! = 0.
e.g.
! = 3! + 1
The y-intercept is 0,1 because when you plug in 0 for ! you get ! = 1.
e.g.
! =!−4
The y-intercept is 0, −4 because when you plug in 0 for ! you get ! = −4.
e.g.
! = 3!
The y-intercept is 0, 0 because when you plug in 0 for ! you get ! = 0.
Lesson 1.20
Part I: Simplifying - Exponent of a monomial
In lesson 1.18 we learned about adding, subtracting, and multiplying monomials. In
the last lesson we learned about raising an expression with an exponent to an
exponent. Now, we will lean about raising a monomial to an exponent.
e.g.
4! ! !
We actually already learned how to do this when we did distribution in lesson 1.15.
Notice how the 4 and the ! ! are multiplying. Therefore, we can distribute the
exponent of 3 to both the 4 and to the ! ! . Notice that we can’t distribute the 3 to the
2 directly, but only to the ! ! term as a whole:
4! ! !
4! ∙ ! ! !
64! !
It is very important to remember that we are not doing 2! , which would give us 8,
but rather ! ! ! , which we learned how to do in the last lesson, lesson 1.19.
Part III: Functions - Fractional slope
Sometimes the slope is a fraction.
e.g.
!
! = !! +2
!
In this example the slope is ! . Whenever the slope is a fraction you want to plug in
! values that are multiples of the denominator so that they will cancel out to give
you integers for y. This makes it easier to graph. So, in this example, you’ll want to
plug in multiples of 3 for !, i.e. 0, 3, 6, 9, etc.
!
!
Ordered pair
0
2
(0,2)
3
3
(3,3)
6
4
(6,4)
9
5
(9,5)
!
When you graph the line you will notice that the function still goes up ! every time
the graph goes to the right 1, which is the definiton of the slope. Another way of
!
thinking about this is like if you walked ! of a mile an hour. After 3 hours you would
have walked 1 mile. After 6 hours you would have walked 2 miles, etc.