14. Traditional Math: Two-Step Equations with Fractions

Seventh Grade

A previous section discussed solving one-step equations with rational coefficients. Students tend to do well on this when given the lesson, but then seem to have difficulty replicating what they did on a subsequent day. It is not unusual for today’s successes to be tomorrow’s failures. The solution is for a constant repetition of these type of problems worked into homework assignments and warm-ups, as well as on quizzes and tests. Unfamiliarity will breed helplessness (and even contempt) while familiarity will breed success.

This is particularly true for two-step equations with fractions. The approach I have used is a carefully scaffolded one. Problems of the type (2/3)x +5/6 = 7/8 are what we address here, and can help in providing a solid base of experience that will serve students well in a future algebra course.

Warm-Ups.  These warm-ups should start the scaffolding process which began in the previous lesson on solving two-step equations with decimals. This lesson builds upon that, beginning with warm-ups.

The warm-ups should also include having students translate an equation stated in English into algebraic symbols as a preview to a later section on word problems.

1. Solve.  0.05x = 0.2  Answer:  5x=20; x = 4

2. 1/3 - 1/5 = ? Answer:  5/15 – 3/15 = 2/15

3. 0.25m - .3 = 0.2  Answer: 25m – 30 = 20; 25m = 50; m = 2

4. Translate to a math expression and solve. Three times a number increased by five is seventeen.  Answer: 3x+5 = 17;  3x = 12; x = 4

5.  Simplify by using distributive rule.  -10(-2 -0.5x)  Answer: 2 + 5x

Basic Method.  I start out by writing an equation on the board and ask if anyone knows how to solve it.

(3/5)x -2/5 = 4/5

Those students who are confident in their abilities with fractions will usually volunteer an answer. The most straightforward way to solve is to add 2/5 to both sides of the equation, resulting in: (3/5)x = 4/5+2/5, which simplifies to (3/5)x = 6/5

“Are we done?” I’ll ask. The answer, of course, is no. We have to isolate the variable. “How do we do that?”

Some students will say “Divide both sides by 3/5” and others will say “Multiply both sides by the reciprocal of 3/5”.  Both are right, and I will say so, reminding students that we discussed this in a previous lesson. The answer is x = 2.

“What did you notice about the denominators in that problem?”  They were the same, of course, which leads to the next problem: “What do we do if we have a problem like this where we have unlike denominators?”

(5/6)x - 1/8 = 1/3

Some students will say we have to find the lowest common denominator, which in this case is 24.  Actually, we only have to find the common denominator for the fractions that we work with in the addition/subtraction step, but I don’t say anything about that yet.  Making equivalent fractions with the denominator of 24 we now have:

(20/24)x – 3/24 = 8/24

“Now we proceed like we did for problems we’ve done before.” I work through the problem with them, adding 3/24 to each side to obtain:

(20/24)x = 11/24

“What is our next step?”

Multiplying both sides by 24/20 results in x = 11/20.

I will then ask, “Could we have solved this problem without changing the denominator of the variable term?”  I then show how it would work if we did so.

(5/6)x = 11/24, yielding the same result.

We then do a few others so they get the hang of it.

(2/9)x -3/2 = 5/6.   “What is the lowest common denominator?”  I will remind them they only have to look at the fractions they will be adding or subtracting. 

Other examples:   (2/3)x + 2/15= 4/5  Answer:  (2/3)x = 12/15-2/15 = 10/15

x = 10/15 ∙ 2/3;  x= ½

2/3 = x/5 + 4   In this problem the format is different and we have x/5 which is the same as (1/5)x.  If they are stuck I will advise that they write it in the format they’re used to which would be (1/5)x +4 = 2/3.   Since the numbers we will be working with are 4 and 2/3, this may give them pause.  “Can you write 4 as 4/1?” is a hint I might give.  The common denominator would then be 3 and the rewritten problem would now look like (1/5)x +12/3 = 2/3;  (1/5)x = -10/3;  x = -50/3 or -16 2/3.

For a Later Lesson. These problems can also be solved by eliminating the fractions by multiplying each term by the lowest common denominator. This method will be presented in a future lesson that will be part of a unit on more solving more complex equations.