Students have learned the basics of solving systems of equations in accelerated seventh grade math. (See here, here, here and here ) Typically, in an algebra 1 course, I offer a brief review. My experience is that students remember the techniques fairly well, although the substitution method still proves difficult for some students.
They are ready to solve more complicated problems with two variables, such as:
Students have had experience solving basic word problems with two variables. Nevertheless, I find that they need additional experience solving word problems with one variable before learning to solve more complex problems using two variables. An example of this is the type of problem I call “hiding in plain sight”. A problem of this type is: “The sum of two numbers is 46. Five times the smaller number is 6 more than twice the larger. Find the numbers.”
When introducing algebra students to solving two variable problems by substitution, I revisit that problem. I review how they solved it with one variable: Let x = the smaller number; then 46-x is the larger number. (That’s the part students continually find difficult). Then the equation is 5x = 6 + 2(46 – x).
Now I show the two variable method in which x = the smaller number and y = the larger number. The equations then become: x + y = 46 and 5x = 6 + 2y. I suggest that we solve for y in the first equation, obtaining y = 46 – x and substitute 46 – x in for y in the second equation:
I point out that this is the same equation using one variable and that they have actually been using the substitution method for some time, but didn’t know it. There is a collective “ah ha” from the class.
When I first started teaching algebra, I tried to teach the two variable method at the beginning of the year, so students would not get confused over the “hiding in plain sight type of problem.” I found that it didn’t work as well as I thought it would. I therefore teach it later, following the sequence in Dolciani’s algebra book and others.
This section does not go into the methods for solving systems of equations. Rather it is focused on two types of word problems and pairs of linear inequalities. The two types of word problems are mixture and wind/current problems. Students have had experience with both of these types, but confined to one variable. Specifically, mixture problems have focused on how much water or other diluting material is needed to reduce a chemical mixture of a given percent, or how much pure chemical must be added to increase the percentage. These are most easily solved using one variable—with two variables the set-up tends to be confusing. Two-variable mixture problems involve combining two types of quantities—chemical mixtures, or different priced tickets to a movie, and so on.
For wind and current problems, students have solved problems in which either the current or wind speed is given and they must find the speed of the object in still water or air—or vice versa. With two variables, students must find both the current/wind speed as well as the speed of the object.
The last topic in this section addresses graphing pairs of linear inequalities. Students graph the combination of linear inequalities such as y > 3x – 6 and 2y < 6x – 12. This is extended to word problems such as: Tom must buy no more than 30 apples and oranges. For example: Apples sell for $2/lb and oranges for $3/lb. His budget is limited to $70. Write the inequalities and graph them.
Although I don’t address them in this book, in this unit I also teach word problems that focus on digits in numbers, age problems and fraction problems. Examples of these follow:
Digit problem: “The sum of the digits of a two-digit numeral is 6. The number with the digits in reverse order is 12 times the original units digit. Find the original number. Answer: Let t = tens digit and u = units digit; t + u = 6; 10u + t = 12u; u=2, t = 4.
Age problem: “Ruth’s father is 7 times as old as Ruth is. One year ago he was 9 times as old as Ruth was. Find Ruth’s present age.” Answer: Let x = Ruth’s age, y = Ruth’s father’s age. y = 7x; y-1 = 9(x-1); x =4, y = 28
Fraction problem: “A fraction has a value of ¾. When 7 is added to its numerator, the resulting fraction equals the reciprocal of the original fraction. Find the original fraction.” Answer: Let x = numerator of the fraction; y = denominator of the fraction; x/y = 3/4; (x+7)/y = 4/3; x = 9, y = 12
Sys 2: Mixture Problems
There are various types of mixture problems. I address problems that involve 1) mixing two items of various costs such as tickets to a game, and food items and 2) two chemical mixtures of different concentrations. These lend themselves to the two variable solution in which one equation represents the number of items (or the total amount of chemical mixture) and the second equation either the value of the items, or the amount of the chemical in the resulting mixture.
I take two days to do this, since the second type is slightly different than the first. Students have solved mixture problems in which a chemical mixture is either diluted by adding a diluent such as water or alcohol, or in which more chemical is added to increase the concentration. This type lends itself to equations with one variable, while problems involving the mixing of two chemicals of different concentrations lend themselves to two variables.
Students have had some previous work solving basic word problems using two variables. Problem 1 requires students to see that the average is the sum of the two numbers divided by 2. For Problem 3 students typically don’t use an equation to find the total amount of money. Problem 4 will seem difficult to students. Possible prompt: “Is there a way we can make the denominator look like the numerator?”
Defining the Equations. “I’m going to give you a problem that will seem difficult but is actually very easy,” I say and am greeted with looks of skepticism. The following problem is projected on the board:
“A candy shop owner mixed candy selling for 75 cents/lb with another type of candy selling for 50 cents/lb to get a mixture to be sold for 60 cents/lb. How many pounds of each of the two candies must he use to get 50 pounds of the 60 cent mixture?”
“We will solve this using two equations. The first equation is the easiest. It describes the total amount of the mixture. So look at the problem, read it through and tell me what the total amount of candy is in the mixture.”
I let them do it and to build suspense announce the number of hands. “One hand up, two hands, now there are three hands.” They are usually the same people so I wander around the room and help those who seem stuck, suggesting they read the last sentence, and I pick someone who doesn’t raise his/her hand often. They announce that the total is 50 pounds.
“Fifty pounds is the total. Write this in your notebooks. We’ll let x equal the number of pounds of 75 cent candy, and y the number of pounds of 50 cent candy. I want you to write an equation that expresses the total number of pounds of candy.” They usually do this fairly quickly.
“Now we move on to the second equation which is the total cost of the mixture of the two types of candy. We have x pounds of candy selling for 75 cents per pound. If I had 2 pounds of it, how would you figure its value?”
Someone will tell me to multiply $0.75 by 2. “And three pounds?” $0.75 times 3. And x pounds?” There is still a hesitancy but here’s where I call on the usual people in order to keep up the pace. Someone says it’s 0.75x.
“And that’s correct. But it will be easier to count this in pennies so we don’t have to deal with decimals. I’m going to write it as 75x. Now how about the value of y pounds of candy at 50 cents/lb?”
They get this one quickly: 50x.
“Now I want to know the value of the entire mixture. Read the last sentence and tell me the value of the entire mixture.”
Someone will tell me with a qualification that I’ve grown used to hearing: “Well, I’m not sure this is right, but…$30?”
“Yes, how did you do it?”
They disclose that they multiplied $0.50 by 60. “Good. Now we’ll write it in pennies, so it would be 3,000. So we have two costs that we’re adding up to total to $30: The cost of candy at 75 cent/lb plus the cost of candy at 50 cents/lb. Now write out the whole equation.”
I’ll walk around the room providing guidance as needed until I see a prevalence of the following equation:
“Now we have two equations. You can solve by elimination, or by substitution. Your choice!” I pause dramatically and announce: “A hint to the wise: substitution will be an easier way to go.”
Again I walk around and students make their way through the problem. In guiding them toward substitution I tell them that if they solve for y in the first equation, it’s an easy substitution in the second equation. They come up with x = 20 lbs and y = 30 lbs.
Homework. Problems are similar to these; I typically assign 6 problems, since they involve both setting up the equations and solving the systems.