In the previous section (Solving One Step Equations by Division and Multiplication), I erroneously included a discussion of how to solve equations in which the coefficient of the variable is a fraction. I realized later that this topic deserved its own day, rather than to be squeezed in to a lesson, causing students to experience cognitive overload and dark feelings towards math. This section addresses only the method for solving such equations.
In the previous lesson, we focused on the method for solving equations in the form ax =b, like 6x= 12 and -5x = 20 in which both sides are divided by the coefficient of the variable. For equations in the form x/m = b, both sides are multiplied by the denominator. For example x/4 = 12 is solved by multiplying both sides by 4.
As discussed previously, these problems are taught using two principles: The division property of equality, and the multiplication property of equality. We treat them as two different types of problems, although later when students have had a course in algebra, they will learn that a/b is equivalent to a ∙(1/b). Thus equations in the form ax=b are really solved using the multiplication property of equality since both sides are multiplied by 1/a.
In teaching algebra and Math 8 classes, I make a point of showing students this. In seventh grade I mention it in terms of what we are actually doing when we solve equations in the form x/m = b; that is, x/m is the same thing as (1/m)∙x. I also mention that x/5 = 20 can be written as (1/5)∙x as discussed in the previous lesson. I mention it to students without dwelling on it and contributing to their cognitive load. It does serve as a segue (for the teacher at least, and some of the students to the next topic which is solving equations like (2/3)x = 16.
Warm-Ups. Problems should include some review as well as the principle that a number multiplied by its reciprocal equals 1.
1. 3/5 x 5/3 = ? Answer: 1, of course!
2. 4/7 x ? = 1 Answer: 7/4
3. 1/7 (y) = 4 Answer: y = 28
4. m/13 = 2 Answer: m = 26
5. 2/3 ÷ 5/6 = ? Answer: 2/3 x 6/5 = 4/5 (or 12/15 if not reduced to lowest terms)
Fractions as Coefficients. In the previous lesson, we went over problems like (1/5)y = 20. When asked how they would solve it, one response is often “Divide both sides by 1/5” which as I said previously is technically correct and results in y = 20∙5. As I discussed in the previous section, I explain that one can also just multiply both sides by the reciprocal of the coefficient.
I will therefore start off with this problem as a review. Upon hearing that we multiply both sides by 5, I make sure to embellish this by reminding them that this is in effect multiplying both sides by the reciprocal of the coefficient.
Moving on, I will ask if anybody can solve the following problem: (2/5)x = 86. If someone connects the dots and says “Multiply both sides by the reciprocal of the coefficient” I will celebrate the evening with champagne and caviar. And it sometimes does happen that way. But usually, someone will say “Divide both sides by 2/5”, which is fine because I can then show 86 ÷ 2/5 is the same as 86 x 5/2, when multiplying both sides by the reciprocal.
I explain that by multiplying the coefficient of the variable by its reciprocal we end up with 1x, which we write as x. There is only one x on the left hand side; we have isolated the variable. Now if we multiply by the reciprocal on the left side, what must we do to the right hand side?” Students know this drill fairly well.
I provide other worked examples and ones on which they work on independently:
Examples:
3/7(x) =24, which when solved becomes x = 24 x 7/3, x = 56.
-2/9(m) = 42 3/5(t) = 4 (Answer: 20/3 or 6 2/3).
I will also include problems like 3/7(x) = 4/5, which results in a fraction multiplication: x = 4/5 x 7/3 = 28/15
Decimals as Coefficients. Equations such as 0.3x = 0.54 and x/0.04 = 500 are solved the same way as equations with integral coefficients. I start off by introducing a problem, like the first one above, and asking how it would be solved. Upon hearing “Divide both sides by 0.3” or something resembling it, I do so on the board, ending up with 0.54 ÷ 0.3. Usually, .sStudents will move the decimal point of divisor and dividend one place to the right resulting in 5.4/3, which is easily solved: 1.8 is the answer.
I will ask them how they can remove the decimals from the equation entirely. The greatest number of decimal places in the equation is two (the number 0.54). If there is no response, I will ask “What number do we multiply the numbers by to remove the decimals? 5, 10, 100?” They will settle upon 100, and I will remind them that numerator and denominator must both be multiplied by 100. The result is 54/30 which may not be that much easier to work with than 5.4/3, but it is paving the way for the next lesson on two-step equations, like (1/2)x + 5/6 = 7/8 in which they multiply each term by the lowest common denominator.
For Students Who are Curious. Problems such as 3/7(x) = 4/5 can be used to demonstrate why fractional division involves inverting the divisor and multiplying. I discuss this in a later unit in the chapter on algebra. For seventh grade, I provide an explanation though I do not hold students accountable for being able to replicate it. I do think it’s appropriate to provide extra credit on a quiz or test for students to make the demonstration.
The demonstration is fairly straightforward. We can express the solution of the equation as a division problem. That is, we divide both sides by 3/7, resulting in x = 4/5 ÷ 3/7.
Now let’s solve it by multiplying both sides by the reciprocal of 3/7, the coefficient of x in the original equation. We obtain x = 4/5 × 7/3. Since 4/5 ÷ 3/7 and 4/5 × 7/3 both equal x they must be equal to each other; therefore 4/5 ÷ 3/7 = 4/5 × 7/3.