Mathematics

7.EE.B4a Transcript

This is Common Core State Standards support video in mathematics. This is standard 7.EE.4a. This standard reads: solve word problems leading to equations of the form p x plus q equals r and p times, open parenthesis, x plus q, close parenthesis, equals r, where p, q, and r are specific rational numbers; solve equations of these forms fluently; compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach.

Let’s start with an example dealing with perimeter, and we will start also using the form p times, parenthesis, x plus q, close parenthesis, equal r. And again, let’s go with an arithmetic approach. What you want students to do, obviously, is to start with some type of visualization. So, in this case, we know that the perimeter of a rectangle will be all four sides added together. We know that the width is 21, so we label in the 21 for the widths. And then we set up some type of scenario, some kind of sentence, and an equation that contains all the information that will be necessary.

In this case, we’ve got, I’ve got two symbols to represent the widths and two symbols to represent the lengths, equal to a total of 78. So, we fill in the values that we do know, 21 and 21, plus two unknown lengths equal to 78. So, we know that the 21 and the 21 is 42. Now, just approaching this using calculation and some common sense, 42 plus what is 78? Just using, again some calculation, I need a 6 to get me the 8, and then I need 3 more to get the 7. So, yes, that checks out. That would give me my 78. So, I need 36 for the lengths. But, that’s two lengths put together, so, I need to split that up evenly. Thirty-six split two ways would be 18. So, I know that my length is 18, and then I fill in, and then you can go back and actually add all four together and make sure, again, that it is 78.

One thing to watch out for here with your students, when they draw some type of visual representation, like in this case, a rectangle. In this case, the values turn out fairly close. But you might end up with a problem where what the students get for a solution isn’t drawn to scale with their diagram. But, don’t worry about that because, again, the diagram is there just to help in the thinking process. Of course, it’s impossible to know if the scale is correct because you don’t know what the answers are yet.

Let’s take the same problem and approach it algebraically. Again, we’re going to do some of the same basic kinds of thinking. We want a diagram labeled with our width. We have the same kind of thinking as far as, we’re going to have two lengths and two widths added together to be the 78. But, in this case, it will be more formal. We’ll introduce l as a variable for the length and w as a variable for the width. Now, this corresponds back to this form where we have p times parenthesis x plus q equals r. Again, the difference here is that, rather than x and q for my variables in the parenthesis, I have l and w.

And from this point, the solution is very similar to what we did from an arithmetic approach. In this case, we know that the length is unknown, the width was 21, and now, using our distributive property, we would have 2l, plus 2 times 21 is 42 equals the perimeter, which we do know is 78. And then, from here, it’s the same as we did a while ago. Subtract 42 from both sides. We get 36. Divide by 2, and we get that the length is 18 as before.

Let’s try a second example. Let’s approach this algebraically, and this one is in the form p x plus q equals r. The problem reads: you have a sales job that pays you $100 per week plus 20% commission on your sales that week. How much would you have to sell that week for your gross pay to be $500? Well, first we have to use some logic and set up the situation, the relationships. In this case, the money that you get in commissions plus the $100 that you get regardless would be your weekly pay. Setting this up algebraically, my commission is 20%, so it would be .20S being the value of what I sold that week, plus $100 equals p for the weekly pay.

Now, I do have some additional information. I do know that the pay that I want for that week is $500. So, now, it’s just a matter of doing this algebraically and solving. So, we start off by subtracting $100 from both sides, which would give us 400. And now we need to divide both sides by .20. So, doing this manually or with a calculator, we should get that the amount of weekly sales that we need is $2,000. Then, of course, the students go back and double-check and substitute and make sure that their answer is reasonable and that it is correct.

Now, let’s suppose that our students are not as far along algebraically, and we want to approach this same problem arithmetically just using some common sense and the skills and the prerequisite knowledge that students already have. Again, the same type of logic; we know that the money that we get in commissions plus $100 would be our weekly pay. We know that we want $500, and that our weekly sales, 20% of that, would be what we get paid.

Now, just looking at this, well, what plus 100 is 500? Well, it’s 400, so I know that I’m going to get paid $400 in commission. But, that’s only 20% of my sales, so now what do I do? It’s always a good idea to present or use some type of visual. In this case, let’s use a rectangle, a long skinny one, and we split it up 1, 2, 3, 4, 5 ways. The reason for splitting it up five ways is, well, let’s see. What does 20% mean? Twenty percent means 20 out of 100. Now, in simpler form, that would be 1 over 5.

Now, how does this 1 or 5 relate to this context? Well, let’s stop and think. What this is saying is, that for every $5 that I have in sales, I’m going to get paid $1 out of those 5. Now, if I match that up to my diagram, this means that, okay, here’s my total sales. And out of that, 1 of those 5 will be what I get paid, and I already figured out in my head that, well, hey, I need $400 plus the $100 to get the $500 for the week.

Well, by definition, all of these smaller pieces have got to be the same. So, I simply have 5 of these same 400s throughout, and 5 times 400 is $2,000, which is what we got a while ago using an algebraic approach. Now, in looking at the thinking that was used in this problem using the arithmetic approach, there really was a lot of thinking involved. A lot of times using an algebraic approach, students pretty much just figure out a formula or an equation, and then just plug in the numbers and do the computation. But, notice here, without some type of formal equation that, with variables and everything, that there was a lot of thinking involved. And, in fact, there was a lot of proportional reasoning involved here, which is very important for your middle school students. They have to transition over and really get a lot of experience with proportional reasoning. So, there is a lot of logic to this standard in having students approach problems from both an arithmetic and an algebraic approach.

Let’s try one more example, and this time let’s use the form p times parenthesis, x plus q, close parenthesis, equals r. Let’s start with an arithmetic approach. The problem reads: you have the same number of 20-dollar bills and 100-dollar bills for a total of $960. How many bills of each do you have? Now, let’s suppose that some students switch over and their thinking goes to looking at the monetary value. So, if we look at the monetary value here, your 100-dollar bills and your 20-dollar bills, when you compare the two, the 100-dollar bill is five times larger than the 20-dollar bill. So, what you have here, if we were to draw, a diagram as a, again, a long rectangle split up into six pieces where the first five represent the value of the 100-dollar bill, which would be 5 times the one that’s left. So, it’s a 5 to 1 ratio. Now, looking at this in total, that is six parts. So, if we took the $960 and split it up six ways, again, doing the math very quickly, that would give us a total of 160. Now, that means that each one of these smaller parts is $160.

Now, applying this back to the context of the problem, if I have $160, how many 20-dollar bills would that be? Well, let’s see, 20 into 160, that’s 8. So, I should have 8 20-dollar bills to get the 160 for that rectangle. And then, for the other five rectangles, it’s the same situation. Each one of those is worth 160, and that’s in the form of eight bills, and these are 100-dollar bills over here. Does that check out? Well, let’s see, 8 times 100, that is $800. Added back to the 160, from the 20-dollar bills, yes, that checks out. That is a total of $960.

Now, all students aren’t going to think the same way. Some of them might approach this thinking more along the lines of the number of bills, and we know that the number of bills is the same for the 100-dollar and the 20-dollar bills. Well, let’s see. This would be my 100-dollar bills here, and this would be my 20-dollars, my 20-dollar bills over there, again, using that same rectangular diagram that we did with the first approach. Now, there isn’t any such thing, but in looking at this, if we look at this holistically, that’s $120 total. There is no such thing as a 120-dollar bill, but let’s pretend that there was because that’s what the thinking sort of parallels here.

So, if I had some number of 120-dollar bills, how much would that be if I had $960? How many bills would I have? So, if I divide 120 into 960, I get 8. So, 8 times 120 is the 960, but of course, I have to split that up to where, I don’t really have 120-dollar bills. I have so many of 100, and so many of 20. I said that the number was 8, and it does check out as before. Eight bills that are 100, eight other bills that are 20, and that would give me my total of 960.

If we look at this same scenario from an algebraic approach, we’ve already actually done a lot of the thinking and the solution to this using the arithmetic approach. In fact, it was the second one, where some of the kids were thinking in terms of 120-dollar bills. If we take this form of the equation, and substitute variables that match this context, let’s let b be the number of bills, h be the value of the 100-dollar bills, t be the value of the 20-dollar bills for my $960 total. And so, what we would have here would be so many bills, the 100-dollar bill is worth 100 dollars and the 20-dollar bill, of course, is worth 20, and we end up with something that was very similar a while ago because we would add the 100 and the 20. That would be a 120 times the number of bills equals to 960, which is again, what we just did a while ago.

Most of the time though, rather than start off in this form, sometimes logically, it makes more sense to use the distributive property and start off in this form. And that’s really what happens here also because what would turn out to be here would have been—I have so many 100-dollar bills, plus so many 20-dollar bills to be my $960. So, again, the thinking is very similar. The algebraic approach is a little bit more formal. But again, hopefully, we see the value of both approaches, algebraic and arithmetic, because the thinking is quite the same, although again, there is a lot of richness in approaching this arithmetically where students use a lot more of their previous experience and their other knowledge in order to put the pieces together.