LE 12: Traditional Math: Solving Word Problems Using Two Variable
Accelerated seventh grade and Math 8
An earlier section addressed solving word problems including those with more than one unknown, but using one variable. These type of problems have been part of the Warm-Ups during the past several lessons on solving systems of equations. This has been done in anticipation of this lesson in which students learn how to solve word problems using two variables.
An example of a problem is: “Daisy and Ed sold magazines. Daisy sold $30 more in magazines than Ed. Their combined sales were $330. Find each person’s sales.”
Working with one unknown, a solution would be to let x equal the amount of Ed’s sales, and x+30 equal the amount of Daisy’s sales. (Another would be to let x be Daisy’s sales and then x – 30 would be Ed’s sales, but most students tend to do the problem the first way.) Then the equation would be x + x + 30 = 330 which is easily solved.
With two variables, the amount of sales can be represented by x (Ed’s sales) and y (Daisy’s sales). Then x + y = 330 and y =x+ 30. This systems lends itself to the substitution method. In fact, when substituting x + 30 for y in the first equation, the result is the same equation when using one variable: x + x + 30 = 330.
This raises the question of whether it might be advantageous to teach working with two variables earlier. From what I’ve seen, seventh graders (as well as students in an algebra course) have difficulties with word problems whether one or two variables are used. I taught an algebra course in which the textbook introduced systems of equations in the second chapter. Teaching systems of equations earlier didn’t seem to improve students’ proficiency at solving word problems. If anything, they were confused about working with two variables. It became obvious that they needed more experience working with one variable equations. I think it’s even more so for seventh graders who are encountering equations for the first time.
The question of whether one or two variables should be used in solving a word problem is worth addressing, but I do so in an algebra course. Students will have had more experience and are more responsive to such information.
Warm-Ups.
1. Solve using substitution. y = x + 3; y = 2x – 6 Answer: x + 3 = 2x – 6; x = 9, y = 12
2. Mary bought a dress at a discount of 25%. If the amount of the discount was $20, what was the original amount of the dress?
Answer: 25/100 = 20/x; 0.25/1 = 20/x; 0.25x = 20; x = $80
3. A coat costs 3 times as much as a pair of pants. Together they cost $180. How much does each cost? Answer: Let x = cost of pair of pants; 3x = cost of coat. 3x + x = 80; 4x = 80; x = $20 (pants), 3x = $60 (coat)
4. Solve. 3x – 2y = 15; 7x -3y = 15; Answer: Multiply 1st eqn by -3 and 2nd eqn by 2: -9x + 6y = -45; 14x -6y = 30; 5x = -15; x = -3 y = -12
5. Cindy is 2 years older than Marie. Their ages total 30. How old is each person? Answer: Let x = Marie’s age; x + 2 = Cindy’s age; x + x + 2 = 30; 2x = 28; x = 14, x + 2 = 16.
The word problems in the Warm-Ups will be used in today’s lesson to show how each of them can be solved using two variables. Problem 1 may not be obvious at first. Prompts: “Can you substitute the right hand side of the second equation for y in the first?” Another might be a statement of the transitive property: “Two expressions equal to the same number are equal to each other.” Problem 2 is a review of a discount problem; students may need to be prompted as to how to set up the proportion. “Is the amount of the discount is different than the discounted amount?” “What does the 25% represent?” Problem 4 requires each equation to be multiplied by a number to cancel one set of variables.
The Wind-Up and the Pitch: Solving Equations With Two Variables. The day’s Warm-Ups intentionally contain word problems, as has been the case for the last few days. The purpose is to re-familiarize students with the technique used to solve them. I use the Warm-Up problems as the entrée to the day’s lesson.
Having finished the Warm-Ups and answered questions on the previous day’s homework, I then turn to one of the problems. “We’ve been solving word problems with one variable although the problems may ask for more than one unknown. So today, we’re going to use two variables to solve problems that ask or two unknowns. Let’s look at Problem 5 again: ‘Cindy is 2 years older than Marie. Their ages total 30. How old is each person?’
“Let’s let x represent Cindy’s age and y Marie’s age. How would I translate the first sentence in that problem into an equation using the two variables?” I have them write it in their notebooks or on mini-whiteboards as I walk around the class, offering suggestions as necessary. I will see some equations as x- y= 2, and others as x = y + 2. Sometimes there is a third but not often: x – 2 = y. I write the first two equations on the board. “These are really the same equation are they not? But let’s leave them both as is; both are correct. Now let’s translate the second sentence.” Again I walk around checking what they have. There is one equation for this one. I write the equations as two sets of systems.
“So how would we solve the first system? Substitution or elimination?” The vote comes in as substitution and I have those students who wrote that equation solve it that way in their notebooks.
“What about the second system?” The vote is elimination and those students go to work on that one. The results are mostly the same with the exception of errors: x = 16, y = 14.
The next is one they’ve seen before: “John has $100 more than his sister. Together they have $110. How much does each person have?”
“How would you set this one up? Talk it over; I’ll give you one minute.” At the end of the minute, I’ll ask if anyone has come up with two equations. I’ll get a few people who volunteer, though there are likely more who are afraid of risking being wrong.
John’s money is represented by x, and his sister’s by y. The consensus is to solve it by substitution, and they arrive at x = $105, y = $5, as they did the first time around.
Examples. I continue in this fashion, having the students write in their notebooks as I answer questions for each problem.
1. The Kickers beat the Astros by 3 goals. There was a total of 13 goals scored in the game. How many goals did each team score? Answer: x = no. of Kickers’ goals; y = no. of Astros’ goals; x + y = 13; x – y = 3; x = 8, y =5
2. The sum of two numbers is 35. Their difference is 13. Find the numbers. Answer: x = 1st no., y = 2nd no.; x + y = 35, x – y = 13; x = 24, y = 11
3. The Outing Club has 35 members. There are 3 more climbers than there are skiers. How many climbers are there. Answer: x = no. of climbers, y = no. of skiers; x – y = 3, x + y = 35; x = 19 climbers.
4. The length of a rectangle is three times the width. The perimeter is 48 cm. Find the length and the width.
For this, I work with the class, first asking what the formula is for the perimeter of a rectangle.They have covered that in sixth grade, but in my experience they tend not to remember it. I write it down: Perimeter = 2L + 2W, where L = length, and W = width. First I have them write an equation for the first sentence. It is L = 3W.
The second sentence I have them work in their notebooks and offer guidance as I walk around. The most common mistake is not multiplying each dimension by two, even though the formula is on the board. A common prompt is “What does the formula say? Where are the ‘two’s’ in your equation? Where should they be?” The second equation is 2L + 2W = 48.
Using substitution: 2(3W) + 2W = 48; 6W + 2W = 48; 8W = 48; W = 6, L = 18
Homework. The homework are similar problems. I have used problems from Dolciani’s “Basic Algebra” but there are other algebra books that provide straightforward problems.
An additional lesson in word problems is on cost problems where a typical problem is: “Jack buys 2 CDs and a DVD for $21.15. Nancy buys 3 CDs and 2 DVDs for $36.00. How much does each item cost?” These are a bit different than the ones presented here and should be in a separate lesson.