Word problems are admittedly hard for most students. It requires being able to use algebraic symbols to represent what is going on in the problem. In the various warm-up exercises that lead up to this particular lesson, students will have solved problems that require translating English into algebraic symbols. Thus, they have had some experience with equations—likely without realizing it.
The key is to start with simple numeric problems that are fairly straightforward to translate, and then progress to more complex problems. One difficulty with starting simple is that some of the problems may be solved without resorting to writing an equation. For example, the problem “What number added to 3 is 10?” I have had to tell students that they can easily solve the problem without an equation, but we have to start somewhere. They need to be able to express the problem as x + 3 = 10. Eventually they will encounter problems that do not lend themselves to being solved quite so easily and an equation helps.
In particular, this lesson also includes solving problems with two unknowns. This entails solving equations similar to 3x + x = 8, so they will be learning how to translate words into this form, and then to combine like terms in order to solve.
Adding to the difficulty of solving word problems are issues that come up when solving equations. Students may still be having difficulty combining negative numbers. For example, the equation x +3 = -6 is solved by adding -3 to both sides—otherwise known as subtracting 3 from both sides. The result is x = -6 -3, which is -9. Students tend to get confused with two negatives in a row, so it is good to go over this as the issue comes up. One way is to express the solution as x +3 + (-3) = -6 + (-3) as a reminder that they are “adding the opposite” when subtracting.
Warm-Ups. The warm-ups should include some translations into algebra that are actually part of the lesson they are to learn.
1. Solve. (2/3)x + 3/5 = 7/10 Answer: (2/3)x = 7/10 - 6/10; (2/3)x =1/10; x = 3/20
2. 6 = -5x -4 Answer: -5x = 10; x = -2
3. Translate into an algebraic equation and solve. Five added to two times some number is twenty-five. Answer: 5 + 2x = 25; 2x = 20; x = 10
4. -6 – 2 = ? Answer: -8
5. Translate into an algebraic equation and solve. Some number added to twice that number equals sixty. Answer: x + 2x = 60; 3x = 60; x = 20
One Unknown. Students have already had some experience solving word problems in warm-ups that involve translating numerical statements into equations. For example, problem 3 of the warm ups involves only numbers. They know that “some number” or “a certain number” is to be represented by a variable. They also know how to represent “two times some number” using a variable (2y, 2m, etc).
Numerical problems are good starting points for problems and students are surprised that they are able to do them—perhaps not realizing how much they have had exposure to them in warm-ups.
I give them a few examples to start with, working through the first two with them; for the rest, they work independently, and I require them to write the equation and solve it in their notebooks or on mini-whiteboards. Simply writing the answer will not suffice.
A certain number plus 8 equals 15. Find the number. Answer: x + 8 = 15; x = 7
(There will be some complaints about why they need to write an equation for this. The next problem may address such concern.)
Two more than 3 times a number is 8. Find the number. Answer: 2 +3x = 8; x =2
Eight less than three times a number is -23. Answer: 3x – 8 = -23; 3x = -15; x =-5
(There are apt to be mistakes with this one, since students will tend to write 8 – 3x rather than 3x – 8)
Ten increased by 1/6 of a number is 5. Answer: 10 + (1/6)m = 5; (1/6)m = -5; m = -30.
In summary, the following steps are followed:
1. Read the problem to determine what value is unknown in the problem.
(Example: in the last problem given above, the phrase “1/6 of a number” is the unknown quantity)
2. Define a variable to represent the unknown quantity. (Use any letter to represent the number; say “y”; then 1/6 of a number is (1/6)y)
3. Translate into an equation.
More than One Unknown. Problem 5 of the warm-ups presents a different challenge to students who may not be accustomed to seeing “some number” added to “twice that number”. Students need to see that the number in each of the phrases is the same and represented by the same variable; i.e., x and 2x. It’s helpful to give them some practice doing this, so I use a few examples.
Six times a number added to the number. Answer: 6x + x, which is 7x
Five more than a number added to twice the number. (This one might take some scaffolding. I ask “How do we represent ‘five more than a number’ ?” Then “How do we represent twice the number”.
I have them put it together: 5 + x + 2x which is 5 + 3x
The sum of one number and five times that number: x + 5x, which is 6x
A few more as necessary, and then onto solving problems that have more than one unknown. Problem 5 of the warm-ups can be re-stated so that it must be solved for two numbers: “The sum of a number and twice that number equals 60. What are the two numbers?”
If we solve for the variable x, we have identified one of the two numbers. Since x equals 20, then twice that number is 40. Those are the two numbers: 20 and 40. As easy as this sounds, it takes some students some getting used to. The first few examples should therefore be worked through, asking students what each step should be:
“Bob weighs 15 pounds more than Alan. Together they weigh 99 pounds. How much does each boy weigh?”
I will usually take the first step, saying “If I let m be Alan’s weight, how would I represent Bob’s weight?” The answer is 15 + m. Now I will write “Alan’s weight + Bob’s weight = ___” and ask what I should put in the blank. Hearing 98, I write that, and then ask “How do I write this using variables instead of words?” I am looking for m + 15 + m = 99.
The discussion that follows goes something along these lines: “Can we combine terms? Which ones? What does it look like now? Can we solve it?” We end up with 2m + 15 = 99, which I then have them solve, with at least one student remarking that it’s a lot easier to solve the equation than setting it up. Which is profoundly true and I tell them so.
Now that they’ve solved it, they have an answer for what m equals, which is Alan’s weight. “What’s Bob’s weight?” They must add 15. The two weights are 42 and 57 pounds.
“Joe is 2 years older than Ed. The sum of their ages is 28. Find the age of each boy.”
I will have them do the problem in their notebooks or mini-whiteboards and check to see what they’re doing as I go around the room. Since it is the same problem as the first, but with different numbers, they usually repeat the procedure we just did. I am strict about insisting they write an equation, and I also want them to identify variables. That is: x = Ed’s age, x + 2 = Joe’s age.
“A certain number plus twice that number is -24. Find the two numbers.”
Answer: x + 2x = -24; 3x = -24, x = -8.
Now I will give them a problem I told them about at the beginning of the unit on expressions and equations (see Post 4: Algebraic Expressions and Equations). I want them to solve that problem by writing an equation and solving it. The problem was “John has $100 more than his sister. Together they have $110. How much does each person have?”
Letting y = the sister’s amount of money, then 100 + y = John’s amount. The equation will be y + 100 + y = 110. Solving the equation, y =5 and 100 + y = 105.
Homework. I usually make up a worksheet of about fifteen problems. I divide them up into three groups of five problems: one unknown, two unknown, and mixed problems. The third group entails reading the problem through to determine whether it has one or two unknowns. As usual, I start them on the problems in class, and work with them as they progress through them. I spend time going over the solutions with them the next day.
Problems should ultimately be in the format of variables on one side of the equation only. A later section will cover equations with variables on both sides of the equation and other complexities. For now, we keep it relatively simple, which proves fairly complicated for seventh graders the first time around.
(To find appropriate problems that are like those described here and not overly complex, the beginning word problems in algebra textbooks are one source.)