# Is this way of teaching how to solve equations dangerous somehow?

Two years ago, I bought the book Mathematics for the Nonmathematican, by Morris Kline.

There I learned a new way of solving equations, which is related to the principle that states that any alteration on both sides won’t alter the output, Ex:.

$$x+8=4\tag{1}$$

$$x+8-8=4-8\tag{2}$$

$$x=-4\tag{3}$$

In my school, they didn’t teach me this way, they taught me that the numbers walk from one side to another and when they do that, they change their sign and in one side we should put all the terms that have x, in the other, we put all terms that don’t have $x$’s. Ex:.

$$x+4=2x+7\tag{1}$$

$$x-2x=7-4\tag{2}$$

$$-x=7-4\tag{3}$$

$$-x=3\tag{4}$$

If the $x$ is negative, then you swap the $-$ for $+$ and vice-versa for both terms:

$$x=-3\tag{5}$$

Some days ago I watched a video that stated the importance of teaching long division, the author argued that only long division could teach the concept of convergence and then I got worried with this method, is this way of teaching how to solve equations dangerous somehow? When I learned the first method two years ago, I felt that I could handle my equations better but that could be only wishful thinking.

EDIT: For the ones curious about the video in which the guy says that only long division teaches convergence, you can see it here. Specially this comment.

#### Solutions Collecting From Web of "Is this way of teaching how to solve equations dangerous somehow?"

The worst thing to teach is that mathematics is a series of recipes to be blindly followed. Mathematics should be about ideas. The reason “walking from one side to the other” works is precisely that it preserves the solutions to the equation. Teaching the mechanics of solving equations without even mentioning that reason should be a criminal offense.

The textbook from which I learned basic algebra used exactly that idea. Further, I think numbers walking to the other side teaches use without understanding and should be avoided.

I think the way described in Morris Kline’s book is better.

It teaches the basic principle – the meaning of the equal sign =.

In your first example, you add (-8) to both left hand side and right hand side when going from (1) to (2). In other words, you are solving the equation (to find the value of $x$) by keeping the left hand side equal to the right hand side.

I don’t see any basic principle behind walk from one side to another.

Convergence is a much more advanced concept than a simple equation. It involves more than the basic principle of equality. If one fails to comprehend what equality means, I don’t see how he can understand convergence.

So, yes. It is dangerous to teach students how to solve equations without introducing them the basic principle behind it.

I don’t see how teaching the “perform the same operation on both sides” process is unusual from a pedagogical standpoint. This was how I was taught to solve equations, and every math textbook I’ve read teaches it this way.

Remember that when you’re solving a equation, the end goal is to isolate the variable you want to solve for such that the variable by itself appears only in and by itself on one side of the equation, and a simplified expression is left on the other side.

The second method simplifies this process for certain cases. However, it does not convey the end goal of isolating the variable. Teaching mathematical procedures without understanding the end goal is only a recipe for failure.

It is more general (and thus more useful) than walking numbers that change signs, but slightly dangerous because it only holds for certain operations, which the rule you cited fails to mention.

Background

When we solve an equation, we want to know which values for the variables make the equation true. Any derivation step should leave this set of values unchanged (and hopefully make the equation simpler).

Applying the same operation to both sides of an equation will preserve all existing solutions. However, for some operations, it can add new solutions. For instance, if we take the equation

$$x = x + 1$$

and multiply both sides with 0, we get

$$0x = 0(x + 1)$$

Clearly, x = 4 is a solution to the transformed equation, but not the original one.

More subtly, if we take the equation

$$x = 4$$

and multiply both sides with x, we get

$$xx = 4x$$

Now, x = 0 is a solution for the transformed equation, but not the original one.

The correct rule

Applying any invertible operation to both sides of an equation will leave its set of solutions unchanged. An operation is invertible if it has an inverse operation such that: for every value, applying the operation, and then the inverse operation, will yield the original value.

Proof: We have seen that applying an operation to both sides will preserve existing solutions. Any solution to the original equation will be solution to the transformed one. Because the operation is inverse, we can also apply the inverse operation to the transformed equation, yielding the original one. Therefore, any solution of the transformed equation will also be a solution of the original one. Therefore, both equations have the same solutions.

Now, which operations are invertible?

• Adding any number is invertible (the inverse operation is subtracting that same number).
• Also, multiplying by any number other than 0 is invertible (the inverse operation is dividing by that number).
• One might be tempted to think that taking the square root is the inverse operation of squaring a number – and it is, but only if we know the sign of the original number. (9 is the square of 3, but also the square of -3. To undo squaring, we need to know the sign).

A cautionary tale

I fondly remember when our teacher, who was quite overweight, derived:

Let W be my weight, I my ideal weight, and x my surplus weight. We then have:

$$W = I + x$$
$$Wx = (I + x)x$$
$$0 = x^2 + (I-W)x$$
$$\left(\frac{I-W}{2}\right)^2 = x^2 + (I-W)x + \left(\frac{I-W}{2}\right)^2$$
$$\left(\frac{I-W}{2}\right)^2 = \left(x + \frac{I-W}{2}\right)^2$$
$$\frac{I-W}{2} = x + \frac{I-W}{2}$$
$$0 = x$$

So you see, I am not overweight at all!

You should now be able to spot where he cheated.

The first method includes, in some sense, the second method. You can “walk from one side to another”, because any alteration on both sides won’t alter the output. For your example

 x+4=2x+7 implies x-2x=7-4, because given x+4=2x+7 we have
x+4-4=2x+7-4 which implies
x+0=2x+7-4
x=2x+7-4 which implies
x-2x=2x-2x+7-4
x-2x=0+7-4
x-2x=7-4


The second method only applies in certain situations also. If I’ve understood things correctly here, you can only use the second method when you have a group structure. However, you can always alter both sides equally without altering the output, or equivalently alter both sides in the same exact way. So, the first method comes as more general.

So, yes, teaching the second method exclusively is dangerous. It will kill future algebraists (o. k. that’s an exaggeration, and a joke… but the second method doesn’t come as general of a method as the first).

Your “new” way is the way I was taught; you do the same thing to both sides of the equation. It was also explained, through demonstration and a lot of practice, that you can do this to cancel out a term on one side, “bringing it over” to the other side with the opposite sign. The goal of this process is to isolate your variable of interest as being the only thing on one side of the equation with no other instances of it on the other side, thus stating what that variable, in singular form, equals.

In my school, they didn’t teach me this way, they taught me that the numbers walk from one side to another and when they do that, they change their sign and in one side we should put all the terms that have x, in the other, we put all terms that don’t have x’s.

This is… an incomplete way to teach the concept, to say the least. It is a simplification of what’s going on behind the scenes that may get the student to the correct answer, but why it works remains a mystery, and that can be dangerous when you try to apply these simple rules in situations where the underlying principle contradicts them.

If I were teaching the concept of solving equations through balanced operations to a group of 5th/6th graders for the very first time, I’d illustrate it with the most direct analogy I can think of; a balance scale. Whatever you do, you have to keep the scale balanced. You can add and remove weights, you can tie helium balloons onto the scale, and whenever you see that you have a balloon and an equal opposing weight on one side, you can remove them; through all of this, that needle has to stay in the center. If it moves to the left or right, you have unbalanced the scale, and so what you end up with won’t be equal.

As the others have said walking to the other side is a simplification of the basic principle that a change made to one side of the equation has the same effect on the other. Another reason to stay away from walking to the other side is it empowers to correctly handle more complex equations.

For example if you have $4x+8y=16$, someone just thinking of variables as walking may have a harder time correctly isolating the x. Rather than thinking of it as

$$\frac{4x + 8y}{4} = \frac{16}{4}$$
$$\frac{4x}{4} + \frac{8y}{4} = 4$$
$$x + 2y = 4$$

Thinking like that also makes it easier to avoid mistakes. I’m sure I’ve done that with problems involving exponentiation and division.

I was taught to ‘walk numbers from one side to the other’ (although I don’t think they called it that), but I currently use the principle that an alteration on both sides won’t alter the output. Your second example (walking numbers from one side to the other) is to me the same, but with some steps hidden:

$$x+4=2x+7\tag{1}$$

$$x+4-2x=2x-2x+7\tag{1a}$$

$$x+4-4-2x=7-4\tag{1b}$$

$$x-2x=7-4\tag{2}$$

$$-x=7-4\tag{3}$$

$$-x=3\tag{4}$$

$$-x\cdot-1=3\cdot-1\tag{4a}$$

$$x=-3\tag{5}$$

If you know the rules and you can correctly make these hidden steps (change the sign, etc), then there is no problem. Otherwise, I make the steps explicit just to be sure I’m doing it right.