Writing and Evaluating Algebraic
So in this lesson, what we’re gonna
look at first of all is algebraic expressions and how one could be formed or
written. And that might be from a set of
instructions or from a geometry problem or diagram. And then we’re also gonna have a
look at how we evaluate them. So if we have an expression, what
would happen if we substituted in some values? What value would we get for that
So, first of all, what we’re gonna
do is we’re gonna take a look at some algebraic notation. So, as I said, we’re gonna take a
look at algebraic notation first. So what we’re gonna look at is all
the basic notation that’s gonna help us to write our algebraic expressions.
Well, first, we’re gonna have a
look at 𝑚 multiplied by five. Now how would we write this using
algebraic notation? And the way we would write this is
five 𝑚. We don’t need to write the
multiplication sign in between. But do note we would not write it
𝑚 five. So it has to be the number or
numerical value first.
Then we have 𝑚 divided by
five. Could we rewrite this in any other
way? We’d often see this written as 𝑚
over five because this means 𝑚 divided by five. It couldn’t be five over 𝑚 because
this is something different. This means five divided by 𝑚. But in fact, what we’re doing is 𝑚
divided by five.
Now we’re gonna move on to the next
bit of algebraic notation. If we have 𝑚 multiplied by 𝑚,
this is equal to 𝑚 squared. Now a common mistake here would be
to write two 𝑚. But that would be 𝑚 plus 𝑚, would
be two 𝑚, whereas 𝑚 multiplied by 𝑚 is 𝑚 squared because anything multiplied by
itself is that thing squared.
Another thing to bear in mind with
algebraic notation is if we add variables, so, for instance, if we have two 𝑚 plus
𝑚, then this is gonna be equal to three 𝑚. Because we deal with 𝑚 as one 𝑚,
so two plus one, which is three 𝑚. It’s worth noting that you cannot
add unlike terms. So we couldn’t add different
letters or, for instance, we couldn’t add two 𝑚 to an 𝑚 squared. So it’s got to be of the same
order, and it must be the same variable.
So we can now try and form a very
simple algebraic expression. So we’ve got five more than 𝑚. So how would we represent this? Well, we’d write this as 𝑚 plus
five because it’s five more than 𝑚. And then we could try another one,
𝑚 less than five. And this would be five minus
𝑚. Another common mistake we need to
avoid is writing 𝑚 minus five because this is not the same thing. This is five less than 𝑚, but we
want 𝑚 less than five.
Okay, great, so we’ve looked at
algebraic notation. We’ve formed some very simple
algebraic expressions. Now let’s have a look at how we
would evaluate a simple algebraic expression.
Evaluate 𝑚 plus 𝑘, given that 𝑚
equals seven and 𝑘 equals 15.
Well, if we look at the way we’d
evaluate, what this means is that we need to work out the value of 𝑚 plus 𝑘 given
that we know that 𝑚 equals seven and 𝑘 equals 15. So what we’re gonna do is we’re
gonna substitute 𝑚 equals seven and 𝑘 equals 15 into 𝑚 plus 𝑘. So what we’re gonna get is seven
plus 15. Well, this is gonna give us the
value 22. So therefore, we can say that 𝑚
plus 𝑘, given that 𝑚 equals seven and 𝑘 equals 17, is equal to 22.
Okay, so great, we’ve had a look at
algebraic notation. And we’ve seen how to write very
simple algebraic expressions. And we’ve also evaluated a simple
algebraic expression. So we’ve done the things that we’ve
set out to do for this lesson. However, there is a lot more to
this. So now let’s consider how we might
create an algebraic expression from a real-life context.
A club is producing posters. If the cost of producing one poster
is 10 Egyptian pounds and the shipping cost of an order of any size is 27 Egyptian
pounds, what is the total cost of producing and shipping an order of 𝑥 posters?
So what we’re gonna do with this
question is form an algebraic expression. And we’ve got some information here
to help us. So we know that the cost of
producing one poster is 10 Egyptian pounds. The shipping cost of any size is 27
Egyptian pounds. And what we’re trying to find the
total cost of producing and shipping is an order of 𝑥 posters.
Well, our production cost is gonna
be equal to 10, because it’s 10 Egyptian pounds per poster, multiplied by the number
of posters. Well, we’ve already said that is
𝑥. And using our algebraic notation,
we’re gonna write this as 10𝑥. And our shipping cost is 27
Egyptian pounds because this doesn’t change with the number of posters that there
are. And it’s worth noting that these
are both Egyptian pounds, so 10𝑥 Egyptian pounds and 27 Egyptian pounds.
So therefore, if we want to produce
an expression for the whole lot, it’s gonna be 10𝑥 plus 27 Egyptian pounds. So this is the total cost of
producing and shipping an order of 𝑥 posters.
So great, we’ve now written
expressions, we’ve evaluated simple expressions, and we’ve also looked at real-life
context with our expressions. But what we haven’t looked at is
what we’d do if we wanted to create an expression and then evaluate that expression
in a real-life context if it had more than one variable. So let’s get it on and have a look
at an example of that now.
Using 𝑥 to represent the cost of a
notebook and 𝑦 to represent the cost of a folder, write an expression for the cost
of three notebooks and two folders. Then find the total cost, given
that the notebooks are one dollar and 69 cents each and the folders are zero dollars
and 59 cents each.
Well, if we’re going to form our
expression, we can see that we’ve got 𝑥, which represents the cost of a
notebook. And we’re gonna have three
notebooks. So we can write three 𝑥. And then we’ve got 𝑦, which gonna
represent our folders. And we’ve got two folders, so we’re
gonna add this. So therefore, our expression for
the cost of three notebooks and two folders is three 𝑥 plus two 𝑦.
Okay, great, so now let’s move on
to the second part of the question cause we want to find the total cost given that
the notebooks are one dollar and 69 cents each and the folders are zero dollars and
59 cents each. So what we’re gonna do is we’re
gonna substitute in some values for our 𝑥 and 𝑦. So we know that our 𝑥 is gonna be
one dollar and 69, and our 𝑦 is gonna be zero dollars and 59 cents. So therefore, to calculate the
total cost, what we’re gonna do is three multiplied by 1.69 plus two multiplied by
0.59, which is gonna give us 5.07 plus 1.18. Which gonna give a total cost for
the three notebooks and two folders of six dollars and 25 cents. Okay, great, so we’ve solved this
problem. And we’ve found our expression and
our total cost.
Okay, great, so we’ve now solved
the problem. But we’ve had to find an expression
and evaluate it when we’ve got more than one variable. So let’s move on to our final two
questions, which are gonna help us show what we’re gonna do if we’re gonna apply
these skills to areas where we have perimeter, area, and volume.
A rectangle’s dimensions are eight
𝑥 plus four length units and 16𝑥 plus six length units. Express the perimeter in terms of
𝑥 and calculate the perimeter when 𝑥 equals one.
So what I’ve done first is drawn a
sketch to help us visualize what’s going on. So we’ve got a rectangle, and we’ve
got length 16𝑥 plus six. And we’ve also got width here of
eight 𝑥 plus four. And because it’s a rectangle, we
know that the opposite sides are equal. So we can add these other values
in. So now we’ve got our four sides all
labeled. So what do we do next?
Well, what we’re asked to do is to
express the perimeter. So let’s think how do we find the
perimeter of our rectangle. Well, the perimeter of any shape is
the distance round the outside. And we have a formula for the
perimeter of a rectangle. And that is 𝑃, our perimeter, is
equal to two multiplied by the length plus the width. Or we can also think about it as
the perimeter is equal to two lengths plus two widths.
So the perimeter of our shape is
going to be equal to two multiplied by. Then we’ve got 16𝑥 plus six plus
eight 𝑥 plus four. So we’re gonna get two lots of. And then, first of all, we’ve got
16𝑥 plus eight 𝑥, which is 24𝑥. And then we have positive six and
positive four. So it’s gonna be 24𝑥 plus 10. So great, well, have we finished
Well, no, what we can do now to
simplify even further is to distribute across our parentheses. And when we do that, what we get is
48𝑥 plus 20. And that’s going to be length units
as per the question. Okay, great, so we’ve now found an
expression for our perimeter. So what we need to do is calculate
the perimeter next when 𝑥 is equal to one.
So to find out what the perimeter
is when 𝑥 is equal to one, we’re gonna substitute 𝑥 equal to one into our
expression. So when we do that, we get the
perimeter is equal to 48 multiplied by one plus 20, which gonna give us 68 length
units. So therefore, we’ve solved both
parts because we’ve expressed our perimeter in terms of 𝑥. And we’ve calculated the perimeter
when 𝑥 equals one.
Consider the shown trapezoid. Write an expanded expression for
its area. Simplify the expression, if
There’s also a second part of the
question that we’ll come on to. So what we’re looking at here in
this question is the area of our trapezoid. So we need to know a formula for
this. Well, we know the formula for the
area of a trapezoid. And that is the area is equal to a
half 𝑎 plus 𝑏 multiplied by ℎ. And this is where 𝑎 and 𝑏 are the
parallel sides of our trapezoid and ℎ is the distance between them.
So in our trapezoid, we got 𝑎, 𝑏,
and ℎ, which I’ve labeled. It doesn’t matter which one, the
top or the bottom, is 𝑎 or 𝑏. These are interchangeable. So when we put it together, we’re
gonna get 𝑎 is equal to a half multiplied by 𝑎 minus five plus 𝑎 plus three
multiplied by 𝑎 minus four, which is gonna give us a half multiplied by two 𝑎
minus two multiplied by 𝑎 minus four. So what we could do here is we
could distribute across the parentheses and then half the result. But what I’m gonna start with is by
halfing the first parentheses. And then we’re gonna distribute
across our parentheses.
So when I do that, I’m gonna get 𝑎
minus one multiplied by 𝑎 minus four. Well, if we check what the question
wants, the question wants us to expand the expression and then simplify it. So first of all, we’re gonna expand
or we’re going to distribute across our parentheses. And when we do this, what we’re
gonna get is 𝑎 multiplied by 𝑎, which is 𝑎 squared. 𝑎 multiplied by negative four is
negative four 𝑎. Negative one multiplied by 𝑎 is
negative 𝑎. And negative one multiplied by
negative four, which is positive four.
So now, if we collect our like
terms, what we’re gonna be left with is 𝑎 squared minus five 𝑎 plus four since
this is our expanded expression for the area of our trapezoid. So now what we’re gonna do is we’re
gonna move on to the second part of the question.
So for the second part of the
question, we’re told that the given trapezoid is a cross section of the given prism
of length two 𝑎 minus five. What we need to do is write an
expanded expression for the volume of the prism, then simplify the expression, if
Well, we have the formula for the
prism. And that formula is that the volume
is equal to the cross-sectional area or the area of the cross section multiplied by
the length. Well, we know that the area of the
cross section is equal to 𝑎 squared minus five 𝑎 plus four cause we’ve found that
in the first part of the question. And we’re told in the second part
of the question that the length is equal to two 𝑎 minus five. So therefore, we know the volume is
gonna be equal to 𝑎 squared minus five 𝑎 plus four multiplied by two 𝑎 minus
Now if we distribute across our
parentheses, first of all, we’re gonna have 𝑎 squared multiplied by two 𝑎, which
is two 𝑎 cubed, and 𝑎 squared multiplied by negative five, which is negative five
𝑎 squared. So then we’re gonna get negative
10𝑎 squared plus 25𝑎. Then finally, we’re gonna get
positive eight 𝑎 minus 20.
Okay, now what we need to do is
simplify the expression. And we do that by collecting our
like terms. So then when we’ve done that, we
get that the volume is equal to two 𝑎 cubed minus 15𝑎 squared plus 33𝑎 minus
20. So there we’ve found the area of
the cross section and the volume, and we’ve simplified them to these two
Okay, great, so we’ve reached the
end of the lesson there. Because we’ve moved through the
different stages and shown a number of different problems that involve us creating
expressions and then evaluating these as well. So now the final thing we need to
do is just look at the key points from the lesson.
So the first thing we looked at was
how we used our algebraic notation. So, for instance, 𝑚 multiplied by
five is written five 𝑚, 𝑚 divided by five is equal to 𝑚 over five, and 𝑚
multiplied by 𝑚 is equal to 𝑚 squared. So this is how we use our notation
to represent these. And we also talked about how 𝑚
plus five would be five more than 𝑚 and 𝑚 minus five would be five less than
And then we looked at what
“evaluate” means. And what this means is that we
substitute in a value. So, for example, if we had 𝑚 plus
𝑘 as our expression, then we want to know what happens when 𝑚 is equal to seven
and 𝑘 is equal to 15, then we put these values in. So we’d have seven plus 15, which
will be equal to 22. So we evaluated that expression for
those given values of 𝑚 and 𝑘.
And the final key point is that
we’ve seen that different letters can be used to represent a variety of variables in
different contexts. So it doesn’t matter if we use 𝑥,
𝑦, 𝑚, 𝑎, 𝑏. But we can use these to actually
represent lots of real-life context variables. And we’ve shown that with a number
of examples that we’ve had.