So far we have covered basic operations with algebraic expressions. In most textbooks there are a few more topics that are included in this category—namely addition and subtraction of linear expressions--but my feeling is that these would be better presented in a later unit. At that time, more complicated equations would be covered, namely equations in which variables are on both sides, and multi-step equations. Based on my experiences as well as those of other teachers with whom I’ve worked, students need some time after learning new foundational skills before they take on more complex ones. During that time, while learning the rudiments of solving simple equations, they will also continue to practice the skills they have just learned during their warm-up questions, so that the foundational skills are established. Having those skills mastered helps when learning new skills that build on them. Otherwise, students are left to balance a long list of partially mastered skills that tend to blend in to one another, resulting in students forgetting how to simplify expressions such as (-2)(3x), and other basics.
Warm-Ups for this Lesson. The following type of questions should be covered in the warm-ups preceding the day’s lesson:
1. Simplify by distributing. 2(x – 5) Answer: 2x – 10
2. Simplify. – (–3 – x) Answer 3 + x
3. Combine like terms. –x + 5x +6y -7y Answer: 4x - y
4. Solve for n. n -9 = 5. Answer: n = 14
5. 5 + ? = 0 Answer: -5
Arithmetic Method for Solving Equations. In going through the warm-ups, problems 4 and 5 are germane to today’s lesson. Problem 4 of the warm-ups above is something they have seen since the early grades, and they have solved it using arithmetic methods. I ask how students have solved it. Some may know it using “number bond” relationships. If n-9 = 5, then 5 + 9 = 14, so n = 14. Other students may count up five numbers from 9 to reach 14, and still others may use a guess and check method.
For problem 5, students who remember that the sum of opposite numbers is always zero will have no problem with it. There may be others who have forgotten about opposites. This is a key part of the day’s lesson, so it is good to go over this by giving other examples.
I explain that the sum of opposites will be key in solving equations like the one in problem 4 using algebra, rather than the arithmetic methods that they have used.
Addition and Subtraction Properties of Equality. These are basic rules of algebra, which simply stated is “If the same number is added to equal numbers, the sums are equal” and “If the same number is subtracted from equal numbers, the sums are equal.”
Stated formally, the above is expressed as follows:
For each number a, b, and c, if a = b, then a + c = b + c
To show how this works, I start with an obvious example:
Example: $6,000 = $6,000
$6,000 + $500 = $6,000 + $500
$6,500 = $6,500
In this case, a = 6,000 and b = 6,000
Now, I show a similar example, but this time instead of the form a = b, we have b = x + y. Specifically I use these numbers: a = 50 and b = 35 + 15. Does a = b? Yes, but the initial equation is:
50 = 35 + 15
Add 10 to both sides
50 + 10 = 35 + 10 + 15
I’ll have students try it so they will see the result is 60 = 60. I want to show that the principle holds no matter how the equality on either side is expressed.
Another example:
50 – 10 = 60 – 20
I will ask the students for a number to add to both sides. If they pick “7”, the result is:
50 – 10 + 7 = 60 – 20 + 7
47 = 47
For subtraction, the same principle applies. The formal rule is:
For each number a, b, and c, if a = b, then a - c = b - c
60 = 60
60 – 3 = 60 – 3
57 = 57
And similarly:
75 = 50 + 25
75 -12 = 50 + 25 – 12
63 = 63
Solving equations using these properties. With the above as background, we proceed to solve simple one-step equations using addition and subtraction. These are in the form ax + b = c, where a = 1.
I’ll write an equation on the board and say that now we are going to solve it using one of the properties we just learned:
x + 7 = 25
We want to add a number on the left hand side that will eliminate the 7 and leave x all by itself. This is known as “isolating the variable”. I refer back to the warm-up about adding opposites, and ask “what is the opposite of 7”? Hearing “-7” I then write:
x + 7 = 25
- 7 = -7
I announce that we are going to add – 7 to both sides of the equation. “Whatever we add to one side of an equation we have to add the same amount to the other,” I’ll say. Since 7 + (-7) is 0, we essentially have x + 0 = 25 – 7, which simplifies to x = 18.
I then show them how to check the answer to see if it’s correct. I will sometimes issue the following advice: “On a test or quiz, if you ask me if the answer is correct, I will not tell you. But you can find out by checking it yourself.” Substituting 18 for x in the original equation results in 18 + 7 = 25 which checks.
I’ll have them do two or three more at their desks, writing their work in their notebooks or mini-whiteboards and asking them to check their work.
x + 17 = 19
x + 6 = 2
For this last, they will add 2 + (-6), at which point there may be some confusion. I will remind them that what they are doing when adding a negative number is subtracting. Also, it’s perfectly fine to get a negative number for an answer. In this case the answer is -4. Checking it: -4 + 6 = 2, which is correct.
x+13 = 15 – 14
Although it may seem obvious that in the above problem one just simplifies the right hand side to get 1, and then proceed. But there will be students who will think it is more complicated than that. When met with the question “What do I do?” I ask whether they know what the answer to 15-14 is. They usually do. I then advise them to put that answer in place of the 15-14 on the right hand side. I include this type of problem so they will get used to combining like terms for problems that will require it in future lessons. The next problem is similar.
2x + 5 – x = 13 – 4
Here they will need to combine like terms on both sides to obtain x + 5 = 9.
After doing these examples, I move on to problems with subtraction, in which one needs to add the opposite to both sides.
Examples can continue in the same format as above, but I do like to include a problem for which the answer is zero, to dispel any doubts that zero can be a solution of an equation.
Example: 3x – 2x -5 = -2 – 3
This simplifies to x -5 = -5; 5 is added to both sides to obtain the solution x = 0.
Problems with Fractions. I like to wait until a later time to introduce problems with fractions to ensure students have confidence with their problem solving skills for this type of equation. This will be addressed in a later unit, but for now, I will just say that problems with fractions, like x – 1/3 = 5/6 follow the same principle as above, once a common denominator is found. For the above problem, the equivalent is x – 2/6 = 5/6. Then 2/6 is added to each side to obtain x = 7/6. I include this discussion here as an admonition to try to avoid confusing students at this point but to let you know what you have to look forward to. Sometimes I show a student such a problem and tell them that they will learn how to do such types as well. I usually get either an eye-roll or a “No way!” which is fine.
Homework Problems. The homework problems are more of the same, which students tend to find fun. I like to include additional problems of the type shown here in which they need to combine like terms, but they should be appear after completion of the easier types. Some textbooks like to start with harder problems right off the bat. I always look over the problems and re-order them on a separate worksheet as necessary. The last thing I want is for students to get discouraged before they even begin the assignment.
For the most part, these problems are very easy for students, and that turns out to be the hardest thing—they will want to solve the problems using arithmetic methods. I insist they use the methods shown in the lesson, and that they must show the steps—that is what number is added or subtracted, as well as the intermediate simplification steps when combining like terms is required.
At least one student may ask why the algebraic method is used when what they have done before in early grades also works. My response is that although it may seem like more work, it is a first step in the basics of algebra that is at work in more complex equations where arithmetic methods would be very tedious and time-consuming. I will then show them a problem they will be working on at the end of the unit for the purpose of showing where we’re heading, as well as why we have to start with simple problems.