Lesson Video: Writing and Evaluating Algebraic Expressions | Nagwa Lesson Video: Writing and Evaluating Algebraic Expressions | Nagwa

Lesson Video: Writing and Evaluating Algebraic Expressions Mathematics • 7th Grade

In this video, we will learn how to write algebraic expressions and evaluate them for specific values of their variables.

15:22

Video Transcript

Writing and Evaluating Algebraic Expressions

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 expression?

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 there?

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 possible.

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 possible.

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 five.

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 expressions.

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 𝑚.

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.