### Video Transcript

Letβs take a look at how we evaluate simple algebraic expressions. Algebra is the branch of mathematics that deals with variables. And a variable is a symbol that represents an unknown quantity. But those are all just words. Letβs look at some examples:

π plus three. In this expression the letter π is the
variable. This π represents an unknown quality, an amount that we donβt
know. And π plus three together is an algebraic expression. Itβs algebraic or from algebra because this expression π plus three
contains a variable; it contains π.

And itβs an expression because it contains a number or numbers and at least
one operation. So here we have the operation of addition and the number three to make this an
expression. π plus three is an example of an algebraic expression.

But remember our goal is solving algebraic expressions. So in order to do that,
weβre going to need a little bit more information. Here is our extra information. Evaluate π plus three if π equals five.

My first step here is just to copy down exactly the expression π plus
three. Our next step is to replace the π with a five. After that I add five and three. And I understand that π plus three when π equals five is eight.
So weβve evaluated or solved our expression π plus three with the given information.

Letβs take a look at this example. Solve the expression
below if π equals seven and π equals five.
Our expression is eight plus π minus π.

We just need to start by copying down the expression exactly how itβs listed
in the problem. Then I wanna replace π and π with their
corresponding values, in this case π being equal to five and π being
equal to seven. Now I have an expression thatβs full of numbers and I can follow the order of
operations. Iβll add π and five to give me thirteen. I added the π and five first because in the order of operations we
want to add and subtract from left to right. And finally weβll subtract the seven from the thirteen which equals
six.

When weβre given these values for π and π β when
weβre given π equals seven and π equals five β we can understand that
eight plus π minus π equals six.

Remember how I said earlier that algebraic expressions contain at least one
operation. So far weβve only seen examples of addition and subtraction in
expressions. But expressions can also contain multiplication and division.

In fact hereβs an example of an algebraic expression with
multiplication. Are you curious how this is an example of an expression with
multiplication?
Five π equals five times π.
In algebra the multiplication sign is often omitted. You might see something like nine π ,
three ππ, or even π§π¦. Nine π is the same thing as saying nine times π ,
three ππ equals three times π times π,
and π¦π§ equals π¦ times π§.

Now I want you to take a look at the numbers that Iβve highlighted in
green. In algebra thereβs a special name for these numbers that are being multiplied
by variables. These numbers are called βthe coefficient.β Coefficient is a factor of a multiplication expression.

Hereβs an example of a multiplication expression. Evaluate seven π€ if π€ equals four. First copy down the expression. Next Iβm gonna replace my π€ with the four. And I also added a
multiplication symbol in this time. After that I multiply seven times four.
The solution to this expression is seven times four, which is
twenty-eight.

Hereβs a slightly harder example. Evaluate π¦ squared minus four plus
three if π¦ equals six. Even though this expression has three different operations, we always start
with the same procedure. Iβm sure you guessed βcopy down the expression.β And if you did, you would be
right; thatβs the first thing we need to do. Next weβre gonna replace π¦ with six because that was our given
value.

As you solve more and more complicated expressions, the most important thing
you can remember is that you have to follow the order of operations. We have correctly substituted the six for the π¦. But now what
operation comes first? Since we donβt have any parentheses or brackets in this problem, order of
operations would tell us to go ahead and solve your exponents next. So Iβve solved for six squared which equals thirty-six and copied the rest
of the problem down.

Following order of operations, I now need to add and subtract from left to
right. In this step, I subtracted four from thirty-six. And our next order of operations step will be to add thirty-two plus
three. Our final answer for this expression when π¦ equals six,
π¦ squared minus four plus three equals thirty-five.

Okay, so letβs review. The first thing that weβll do when weβre trying to
evaluate algebraic expressions is to substitute given values for the variables. And after that you need to be very careful to follow the order of operations
to evaluate each operation thatβs in your expression. The best way for you to get good at evaluating expressions is to try it and
then practice. So now itβs your turn.