### Video Transcript

In this video, we will learn how to
add and subtract algebraic expressions by adding and subtracting like terms. The main skill weβll need is the
ability to recognize like terms. Like terms are terms whose variable
and exponent are the same. If we have values like two and 10,
theyβre both whole numbers. So they are considered like terms
and can be combined. However, two π₯ and 10 are not like
terms. In this case, the two has a
variable and the 10 does not. This means these are unlike
terms.

What about two π₯ cubed and 10π₯ to
the fourth power? They do have the same variable. But they do not have the same
variable taken to the same power, which means they cannot be combined and theyβre
not like terms. But if we had two π₯ cubed plus
10π₯ cubed, since both of these terms have the same variable taken to the same
exponent, theyβre like terms and we combine them by adding their coefficients. Two π₯ cubed plus 10π₯ cubed equals
12π₯ cubed. If youβre wondering why this works,
letβs look at this another way.

If we know that a pink block equals
π₯ squared, a yellow block equals π₯, and a green block equals one and we wanted to
rewrite this first set of blocks, instead of saying π₯ squared plus π₯ squared plus
π₯ squared for the pink, we can write that as three π₯ squared. Since there are two yellow blocks,
we have two π₯ and the two green blocks that are whole numbers are two. And if we follow the same procedure
for the second set of blocks, weβll have two π₯ squared plus π₯ plus two. And combining all of these
together, we see that there are five pink blocks, which means there should be five
π₯ squared. And this makes sense because three
π₯ squared plus two π₯ squared would equal five π₯ squared. We combine their coefficients.

When it comes to the π₯-terms, the
yellow blocks, we have three, two π₯ plus π₯ equals three π₯, and of course for our
whole numbers, we have four green blocks, which each represent one, so four
total. In this situation where each term
represents a different color, you would never combine the green and the pink
together. In the same way, we cannot combine
an π₯ squared term with a whole number. Using this skill, we can look at
some examples.

Simplify two π plus one plus three
π plus two.

First, we can go ahead and copy
down our expression. In order to simplify, we first need
to consider if thereβs anything inside the parentheses that can be simplified. Since two π and one are unlike
terms, we canβt add them together. The same thing is true for three π
and two; they canβt be added together. However, because this expression
contains only addition and we can add in any order, we can remove the parentheses to
see if there are any like terms that can be combined. Two π and three π both have the
same variable and therefore can be combined by combining their coefficients. Two π plus three π will be equal
to two plus three π . And we also know that we could add
the two whole numbers together. One plus two is three, which makes
the most simplified form of this expression, five π plus three.

In our next example, weβll first
have to do some substitution. And then weβll be able to add two
expressions together.

Find π΄ plus π΅ given that π΄
equals five π cubed minus three π and π΅ equals negative six π squared plus three
π .

Our job is to find π΄ plus π΅,
which means we plug in what we know π΄ is, five π cubed minus three π . And then we substitute what we know
π΅ is, negative six π squared plus three π . In order to add π΄ plus π΅, weβll
have to see if we can find any like terms. We have an π cubed term, two terms
that just have the variable π , and one term that is π squared. Since weβre only dealing with
addition and subtraction here, weβre able to remove the parentheses. But again, the only values that can
be added together are those that are like terms. That means theyβre the ones with a
variable taken to the same exponent.

In our case, we can combine
negative three π plus three π . We cannot combine five π cubed and
six π squared. Even though they both have the
variable π , their exponents are different. But how do we combine negative
three π plus three π ? Well, negative three π plus three
π equals zero. And that means, in its most
simplified form, π΄ plus π΅ equals five π cubed minus six π squared.

And before we go any further and
look at other examples, we should say that the same rules apply to subtraction of
expressions. You can only subtract like
terms. We could subtract three π₯ minus
two π₯. And we do that by subtracting two
from three, the coefficients, which is one π₯ or just π₯. However, we could not subtract two
π₯ from three π₯ squared. They have the same variable but
different exponents. And so we say that three π₯ squared
minus two π₯ just equals three π₯ squared minus two π₯.

Weβll look at some other cases in
our next example.

Find π΄ minus π΅ given that π΄
equals eight π₯ plus two and π΅ equals five π₯ minus one.

If we want to solve π΄ minus π΅, we
plug in eight π₯ squared plus two for π΄ and five π₯ minus one for π΅. When weβre working with subtraction
of terms, the parentheses are very important. This is because we are subtracting
all of π΅ from π΄. And to do that, we have to
distribute this subtraction across both terms for the expression in π΅. And that means weβre saying eight
π₯ plus two minus five π₯ but plus one because weβre subtracting a negative one. And that means weβre adding. This is the step where if youβre
not careful, you will get a sign mistake.

And so when youβre subtracting
expressions, you need to be very careful to distribute the subtraction
correctly. Once you do that, itβs a simple
matter of combining like terms. We have two terms with an
π₯-variable and two whole numbers. For our π₯-variable, we have eight
π₯ minus five π₯. And that means weβll subtract five
from eight. Remember, the variable doesnβt
change. We just do the subtraction from the
coefficients. And then we have two plus one,
which is three. Eight minus five is three. So we have three π₯ plus three. In this case, under these
conditions, π΄ minus π΅ is equal to three π₯ plus three.

In our next example, weβll deal
with expressions that have three different variables.

Subtract six π₯ to the fifth power
minus three π¦ cubed plus three π§ squared from eight π₯ to the fifth power minus
five π¦ cubed minus two π§ squared.

To solve this problem, we need to
think carefully about how to write out this subtraction. Weβre subtracting six π₯ to the
fifth power minus three π¦ cubed plus three π§ squared from eight π₯ to the fifth
power minus five π¦ cubed minus two π§ squared. And this means that our first
expression will be the eight π₯ to the fifth power minus five π¦ cubed minus two π§
squared. And our second term will be six π₯
to the fifth power minus three π¦ cubed plus three π§ squared. As we are subtracting the entire
second expression from the first expression, we must put that second expression in
parentheses and then distribute the subtraction across every term in the
expression.

In this step, we pay close
attention to the sign of every term. Six π₯ to the fifth power is
positive. And we are going to subtract six π₯
to the fifth power. But the three π¦ cubed is
negative. And if we want to subtract negative
three π¦ cubed, we rewrite that as adding three π¦ cubed. The three π§ squared is
positive. Weβre subtracting positive three π§
squared, which we can rewrite as minus three π§ squared. And in order to do any subtraction
here, we need to see if we have like terms. Those are terms who have variables
taken to the same exponent.

We have one term, that is, π₯ to
the fifth power, and another π₯ to the fifth power term. We then have two terms with the
variable π¦ cubed and two terms with the variable π§ cubed. If you want, you can regroup so
that the like terms are next to each other in the expression. As youβre doing this, pay close
attention to the signs. And then we remember that, to
combine like terms, we combine their coefficients. Eight π₯ to the fifth power minus
six π₯ to the fifth power will be eight minus six π₯ to the fifth power, which is
two π₯ to the fifth power.

For the second terms, we have
negative five π¦ cubed plus three π¦ cubed. That means we need to combine
negative five and positive three, which is negative two. And for our π§ squared term, weβre
combining the coefficients, negative two and negative three. Negative two minus three is
negative five. Putting all this together, we get
two π₯ to the fifth minus two π¦ cubed minus five π§ squared.

Before we leave this question,
letβs look at one other way you could solve, starting with our first expression. Instead of writing the second
expression horizontally, you could vertically line the second expression up
underneath the first. Notice that weβve grouped the π₯ to
the fifth terms, the π¦ cubed terms, and the π§ squared terms. But again, our biggest challenge
here is to make sure we donβt make any sign mistakes. For the first term, weβre saying
eight π₯ to the fifth minus six π₯ to the fifth. So we subtract six from eight. And we get two π₯ to the fifth.

But our second set of like terms
are not as straightforward. We have negative five π¦ cubed. And we are subtracting negative
three π¦ cubed, negative five π¦ cubed plus three π¦ cubed. This is doing that
distribution. Negative five plus three is
negative two. And the variable is π¦ cubed. Then we have negative two π§
squared minus positive three π§ squared, which means weβll have negative two π§
squared minus three π§ squared. Negative two minus three is
negative five. And our variable is π§ squared. And we see here that both methods
will yield the same result.

In our final example, weβll use the
skills of adding algebraic expressions to solve for the perimeter of a right
triangle.

Find the perimeter of the right
triangle shown in the diagram.

We know that the perimeter of any
shape is the distance all the way around the shape. In the case of this triangle, since
all three sides have a different length, we need to add those three sides
together. Side one has a measure of three π₯
plus side two, π₯ plus three, plus side three, two π₯ plus seven. Since weβre only dealing with
addition, we can regroup the terms however we would like. And we want to group our like terms
together. Three of our terms have an
π₯-variable. And the other two terms are whole
numbers.

We could rewrite the expression
grouping the π₯-terms and the whole numbers. From there, for π₯-terms, we can
combine these terms by combining their coefficients, which would be three plus one
plus two, which is six. And to combine the whole numbers,
of course, three plus seven is 10. Since we donβt have any information
about what π₯ is, we can only say that the perimeter of this shape is equal to six
π₯ plus 10.

Before we finish, letβs review the
key points necessary for adding and subtracting algebraic expressions. First, we must recognize like
terms. Like terms are terms whose
variables and exponents are the same. To combine like terms, add or
subtract their coefficients. When adding and subtracting two or
more algebraic expressions, the only values that can be combined are the terms that
are alike.