Equations in the form p(x+q) = r are two-step equations that involve the distributive property. Some textbooks place much importance on equations in which r is a multiple of p, such as 3(x+5) = 18. In this case, both sides can be divided by 3 to result in a simplified form of the equation: x+5 = 6. Although it is nice if students can see this, my experience is that most students will simply distribute and solve. They are like people learning how to drive: eyes fixed straight ahead, focusing on the road and not looking for nuance.
For 3(x+5) = 18, they would distribute the 3 and obtain 3x + 15 = 18.
I’m fine with that and don’t have a preference for how students choose to do it, particularly since there are many equations in which r is not a multiple of p. In those instances the distribution method is the most straightforward. On the other hand, when p is a fraction there is an advantage to dividing both sides by p. For example: 4/5(2+m) = 24 is solved easily by multiplying both sides by the reciprocal of 4/5 to obtain: 2 + m = 24 ∙ 5/4; 2 + m =30. Again, I don’t stress if a student distributes the 4/5. But as I noted elsewhere, students tend not to like fractions, and they quickly gravitate to a method that gives them an edge in the process.
This lesson also addresses word problems that lend themselves to being expressed in the form p(x+q) = r.
Warm-Ups. These include some distribution problems including, and some word problems for which they have received instruction and practice in earlier lessons (See Topic 15: Solving Word Problems Using Equations).
1. Simplify by distribution. −3(𝑥−5) 𝐴𝑛𝑠𝑤𝑒𝑟: −3𝑥+15
2. A bowling alley charges $3 to rent shoes. The cost of a game is $2. If Bob spends $13, how many games has he bowled?
(2/9)∙27 = ? Answer: 6
5. Lois has $48 more than Ed. Together they have $200. How much money does each person have? Answer: Let x = how much money Ed has. Then x + 48 represents Lois’ money. x + 48 + x = 200; 2x =152; x = $76, x + 48 = $124
Problem 1 causes students difficulty, so I rewrite it on the board and provide hints:.
Problem 4 is included because of the tendency for students to forget how to multiply a fraction times a whole number. Many students still need to write 27/1 in order to see that one is just multiplying fractions. My feeling is that if it helps, then go with it. On the other hand, if the problem were “Find 1/6 of 18” it should be obvious that it’s 18/6 or 3.
In Problem 3, for those students who do not elect to multiply all terms by the common denominator, they will need to subtract 4/7 from 6, which entails figuring out how many 7ths equal 6. I’ve seen many student stumble on this, until I write 6/1, and ask what the equivalent fraction is with a denominator of 7.
Solving Equations that Require Distributions. I rarely talk about the objective or goals of the day’s lesson since it becomes fairly obvious once we get into it. When I do talk about goals it’s usually what students will be able to do at the end of the unit. In this case, I usually will mention that we’ve been going over more complex equations and by the end of tomorrow’s lesson they will be able to solve equations like the following:
After hearing a few “no ways”, I then announce that we’ll work on something a bit simpler.
I then write a truncated version of that equation and ask how we should solve it:
Someone will likely suggest that we distribute the left hand side, but if not, I will suggest it. “Let’s see what we get.”
I’ll ask them to solve it which they will do fairly quickly: y=3.
“Now if I had an equation like 5m = 45, what would we do?” It’s obvious that both sides are divided by 5.
“And isn’t the equation we just solved in that same form? We have five multiplied by an expression. How do we undo the multiplication?”
Students will reflexively answer “Divide” which I then do:
Solving it, we get the same answer. “The five outside the parentheses means 5 is multiplied by that expression. So we can undo that multiplication just like we did when we had 5m = 45.”
I try a few others:
I mention that we can also distribute, and I don’t really care which method they use. As I noted earlier, many students choose to use the distribution method.
There are situations when dividing by both sides is not advisable, I will tell them and write an example on the board:
Solving it by distribution is easiest:
You could divide both sides by 4, but it ends up with fractions that most students would rather not have to deal with:
A few more examples, some solved by dividing, and some that can’t then follow so they are comfortable doing these.
Fractional Multipliers. There are problems in which dividing both sides by the multiplier is much more straightforward.
I first remind them of how we solve equations when the coefficient is a fraction:
Students should be able to recall that both sides are multiplied by the reciprocal of 2/3.
The same process is used for the problem on the board as shown here:
A few more examples generally gets the point across.
CONTINUED IN PART II.