Word problems are what students find the most difficult in math. Their difficulty lies in finding out how the information in the problem is to be interpreted and ultimately assembled as an algebraic equation. This requires an identification of what is being asked: i.e., what unknowns are to be identified, and what quantities are being equated.
Like most topics in math, it is best for students to start with simple problems and work their way up to more complexity. Much of this lesson has been addressed in seventh grade, (Post 15) and also in Post AE3. But in this lesson the problems are starting to get a bit more involved beyond the simple number problems such as “A number added to two times the number is 64—what is the number?”
I start with a problem that gets repeated throughout the course, with increasingly complex restatements. In its simplest form it asks to find the length and width of a rectangle given certain pieces of information. (As described in Post AE 3)
Eventually such problems include the concept of area so that by the end of the course, students are having to solve problems like: “Two tin squares together have an area of 325 square inches. One square is 5 inches longer than the other. Find the side of each.”
Math books tend to provide a list of “things to do”. The 1962 Dolciani algebra book is no exception to this, but I find that such lists tend to be regarded as a how-to that applies to all word problems. In her book she lists the steps one should follow but at least she labels them as “four possible steps in solving a problem”:
1. Choose a variable to use in representing each described number in the problem
2. Form an equation by using facts given in the problem
3. Solve the equation
4. Check the answer using the words of the problem
Admittedly these are better than Polya’s problem solving steps that for a time were embedded in many textbooks. Generic problem solving steps are not methods unto themselves—one needs domain content—i.e., the problem itself—to know how to proceed. The first two steps above are in danger of being viewed as a schema or method. They make the most sense in the context of specific problem types. Finding the variable to use may differ depending on what the problem is: rectangle problems have a different structure than distance/speed problems, and mixture problems.
All that said, I have found that the difficulty with word problems persists throughout the course. I view it as an inroad, a kind of muscle-memory of the mind. For me, word problems in Algebra 1 were very difficult. When I was in Algebra 2, the teacher took pains to explain what I have just described and suddenly what I found difficult seemed obvious. I recognize this will not always be the case for all students. There is no royal road that substitutes for hard work and practice.
Warm-Ups.
1. Solve. 5z = -2z +14,021 Answer: 7z = 14,021; z =2,003
2. Solve. 8x -5 = -5 Answer: 8x = 0, x = 0
3. Solve. 2(t + 4) – 3 = ½ (10+6t) Answer: 2t+8-3 = 5 + 3t; 2t+5 = 5 + 3t; t = 0
4. Write an equation and solve. If two times a number is decreased by 16, the result is four. Find the number. Answer: 2x – 16 = 4; 2x =20; x = 10
5. Combine like terms. -2x – 5y – 5x – 6 Answer: -7x – 5y – 6
Problem 3 requires multiplication by one-half. A prompt may be “What is one-half of ten? What is one-half of 6t?”
Expressing one unknown directly in terms of the other. Rather than announce that today’s lesson is on word problems (resulting in groans and general expressions of disdain), I make it obvious that that’s what we’re doing by discussing by projecting the following problem on the board:
“The sum of the length and width of a rectangle is 42 inches. Twice the length is 1 inch less than 3 times the width. Find the dimensions of the rectangle.”
We have discussed rectangle problem as I described in Post AE 3. This one is a little bit more complex. “Who remembers the formula for perimeter of a rectangle?” I hope and pray that someone will and someone usually does. (I make a mental note of who is retaining things like this as a form of ‘checking for understanding’ or formative assessment.)
I write the formula for perimeter of a rectangle on the board: 2L + 2W = P
“Can I start substituting into this formula? What information do I have? Do I know the perimeter?” Yes, it’s 42 inches. Now the formula looks like 2L + 2W = 42
“What should we let x represent? Length or width? Which would be easier?”
The general consensus is width. “So let’s write down the variable to keep track.”
I write x = width of rectangle
“How does the problem define the length? Look at it, and tell me. Raise your hand don’t shout.” (This last phrase has become as automatic for me as I hope the formula for perimeter will be for the class.) I hear “1 inch less than 3 times the width.”
“How would I write length then in terms of width? Where do I start?” If no immediate response, I offer some advice. “Suppose the width were 10 inches. How would I figure out what 1 inch less than 3 times 10 inches is? Do I multiply first or subtract? Does the order of operations, PEMDAS, apply here?”
Someone will say that you multiply 10 times 3 and subtract 1. “Ah ha. So instead of writing 10, let’s write x since that’s the width, and what do we get?”
They should get 3x – 1.
“Now write the equation in your notebooks and I’ll be coming around. Just the equation, that’s all you have to do for now. The easy part is solving it. I want you to do the hard part now.”
I’m looking for 2(3x-1) + 2x = 42. Prompts may include “What do I plug into the formula for L?”
Other examples are found in Post 15. The idea is to get them to find the most likely parameter to be x and then define the other variable(s) in terms of it.
Hiding in Plain Sight. “Suppose now that I have this problem
I project the following problem on the board:
The sum of two numbers is 46. Two times the smaller number plus the larger number is 52. Find the numbers.
There are usually blank stares which tells me they’ve read the problem.
“Let’s write this in English first. That might help. What are we making equal? In other words, what does on one side of the equation, and what goes on the other? HINT: What does the word ‘is’ generally mean?”
At this hint, student’s eyes suddenly light up. “Equals!”
“OK, so let’s write the left hand side:”
Two ∙ (smaller number) + larger number =
“What do I write on the right hand side?” They say “52” and now the equation looks like:
Two ∙ (smaller number) + larger number = 52
“We know from the first sentence in the problem that the sum of larger and smaller is 46. The problem is, if we let x represent the smaller number, let’s say, then how do I represent the larger number?”
Some students may say y, to which I response “I meant in terms of x.” (Not that there’s anything wrong with saying y, but they are not quite ready to work with two variables. I tried that once, and students became very confused. So I like to wait; when we get to the chapter on linear systems, they are ready to take it on.)
“This is what I call a “hiding in plain sight” problem. Suppose the smaller number is 10; how would you find the larger number?” They say “Subtract 10 from 46.”
I draw a diagram on the board:
“That gives me 10 and 36. Suppose the smaller number were five? What would I do?”
They tell me the answer and I update the diagram:
“You get the idea. Now suppose the smaller number is x. If x is the smaller, how do I write the larger number?”
They see the pattern and that it equals 46-x.
“Let’s plug things in. If the smaller number, then how do I writer two times the smaller number?” I hear 2x.
“Let’s add the larger number.”
We now have: 2x + 46-x = 52. “Solve it.” They do and get x = 6.
Example. I admonish students to write down how they are defining the variables.
Two numbers sum to 98. Four more than the larger number is twice the smaller number. This will take some hints. “How do I write the smaller number?” What goes on each side of the equation?” Answer: Let x = smaller number; larger number = 98-x. 98 -x + 4 = 2x; 3x = 102; x = 34, 98 – x = 64
Some may question what would happen if x represents the larger number. In that case, the equation would be: x + 4 = 2(98 – x); x + 4 = 196 – 2x; 3x = 192; x = 64 (the larger number) and 198 – 64 = 34 (the smaller number). Doing this emphasizes the importance of writing down what the variables represent.
Homework. Homework should be a mix of different types of word problems—among them, the “hiding in plain sight” type. I will work one or two of the more difficult homework problems with the class.
For quizzes and tests, I include one or two of the simpler problems in the main part of the exam; more difficult problems are for extra credit. No penalty is given for those; if they get it right, additional points are added to their score.