AE 3 Traditional Math: Solving Word Problems Using Equations
Eighth Grade
Students have had experience with solving equations, as well as word problems from the seventh grade math class. Their experience may vary depending on how their class was conducted. I was fortunate in that I often taught algebra to students who I had had before in seventh grade. Nevertheless, there is some amount of forgetting that goes on and while I wasn’t starting from scratch, I did have to do some review in introducing word problems. In either case, I started with the basics once again, but introduced more problems than I did in seventh grade, which is the approach I will describe here.
The technique and lesson I used for seventh grade is described in Post 15. In this lesson, I refer to that Post for solving problems with one unknown, and focus this discussion on solving for more than one unknown.
Warm-Ups.
1. Trish is years old. How old was she 7 years ago in terms of x? Answer: x - 7
2. Solve. x + 9 = 28 Answer: x = 19
3. Bob is twice as old as Emma. If Emma is x years old, how old is Bob in terms of Answer: 2x
4. A board is 27 inches in length. If the board is cut in two and one of the pieces is 10 inches, find the length of the second piece and write the step used to calculate it. Answer: 27 – 10 = 17
5. Translate into algebra and solve. Seventeen less than 4 times a number is 11. Answer: 4x – 17 = 11; 4x = 28; x = 7
These problems focus on translating words into algebraic symbols. Problems 1 and 2 require expressing one value in terms of x. Problem 3 provides the fundamentals of what is called “hiding in plain sight” which will be discussed in a subsequent lesson. Problem 5 requires translation into algebra, and solving the resulting equation which segues into the day’s lesson.
Using Equations to Solve Problems. As I did in seventh grade, I start off the lesson with the following problem. If my students already had this problem the previous year, I may couch it in terms of “Who remembers how to solve this problem?” which is a handy way to see how much re-teaching I need to do. If I have a totally new class, then I just go ahead and give the problem. With a mix of students—some of whom hadn’t had my previous class—I admonish those who know the answer to keep quiet, even though such admonishment is a bit dicey.
The problem is as follows: John and his sister have $110 between them. John has $100 more than his sister. How much do each of them have?” I have the students solve it. I have described the approach in Post 15 and will not repeat it here, except to say that those who haven’t seen it before use a “guess and check” approach.
I then go into how to solve problems with one unknown, with the key being how to express that unknown in the context of the problem. Again, see Post 15 for a discussion of how I proceed.
Students have difficulty expressing the problem in terms of x when they have been used to solving these same type of problems with arithmetic. They also questioned why anyone would want to solve such problems with algebra. “We have to start somewhere to learn equations, so we start with very simple problems. I promise they will get more difficult.” Students generally don’t like the last part of my answer.
An example of such a problem is “The number of boys in a certain class is seven times the number of girls. The number of boys is 28. How many girls are in the class?” Students will complain that all you have to do is divide 28 by 7. “That’s correct, but I want it expressed as an equation. What is the problem asking for?” It’s asking for the number of girls. Therefore we can let x represent the number of girls. “If x represents the number of girls, how do I represent seven times the number of girls?” 7x. “Do you have enough to write the equation?” In general, there are enough students who catch on, and write 7x = 28.
Examples:
1. Bob is twice as old as Emma. Kent is 3 years older than Bob. If Emma is x years old, in terms of x how old is a) Bob, b) Kent? Answer: a) 2x, b) 3 + 2x
2. Five more than a number is 11. Write an equation and solve. Answer: 5 + x = 11, x = 6
3. Write an equation and solve. After selling 12 cartons of milk, a store had 75 cartons left. How many cartons were there originally? A prompt I use is “What should we let x represent?” The answer is original number of cartons. “How do we represent taking 12 away from that original amount in algebraic symbols?” x – 12. This is generally enough information for them to complete the equation: x – 12 = 75, and x = 86.
More than One Unknown. These type of problems are also discussed in Post 15. The examples I use are similar to those I used in seventh grade. One problem is “The main body of a rocket ship is eight times as long as the nose cone at the top. The entire rocket is 90 feet in length. How long is the nose cone, and how long is the main body of the rocket?”
“There are two unknowns here. What information are we given? Do we know the total length?” Yes, we do; the problem says it is 90 feet.
“What about the length of the main body and the length of the nose cone?” The main body is eight times the length of the nose cone.
I write the following on the board:
Length of cone + length of main body = entire length
“Can we substitute into this? What are we going to have x represent?” The easiest way is to let x represent the length of the nose cone. Then we have:
x + 8x = 90, which is easily solved: 9x = 90, x = 10 ft.
“OK, that’s the length of the nose cone. What’s the length of the main body?” They see it’s eight times that amount, so it’s 80 ft—problem solved.
Examples: (I work out the first one with the students)
1. An 88 seat passenger plane has 10 times as many economy seats as first-class seats. Find the number of first class seats. If is the number of first class seats, then the number of economy seats is 10 times as much, or 10n
First class seats + economy seats = 88; n + 10n = 88; 11n = 88; n = 8, 10n = 80
2. The length of a rectangle is five times its width. The perimeter of the rectangle is 240 ft. Find the length and width of the rectangle. (I remind students of the formula for perimeter of a rectangle: 2L + 2W = P) Answer: Let the width; then length
2 times length + 2 times width = perimeter;
2(5w) + 2w = 240; 10w + 2w = 240; 12w = 240; w = 20 ft., 5w = 100 ft.
Homework. The homework should be a mix of one and two unknowns, and a problem or two can be worked out “for free” when the class starts working on homework problems. As stated in the lesson above, some of the problems are difficult for students because they are used to thinking arithmetically, and setting up the problem to be solved using algebra is admittedly more work. But as I tell students, it is because we need to start somewhere. I have grown to understand and accept that my advice doesn’t satisfy their complaint.