#### (A new question of the week)

Sometimes a problem that looks complicated turns out to have a simple answer. And sometimes that simple answer turns out to be too simple for its own good. Today’s problem is an example of this.

## A nonlinear system

Here is the question, which came in last month:

Sir,

What is the value of c if

a + b = 12

ac + b = 18

bc + a = 6 ?Choices:

a) 7 b) 4 c) 2 d) 1

After our usual reminder that we want to see your own thinking so that we can help you work it out, rather than just give answers, I gave some suggestions for starting:

I can see several ways to solve this by substitution. One somewhat nonstandard way, based on special features of the equations, is to solve the second for b and the third for a, and plug those into the first. You will almost immediately get a value for c.

A more typical approach is to solve the first for b and put that into each of the others. After simplification, the resulting two equations can easily be combined to get c.

What I had seen was that because the last two equations have similar forms, you can solve the second to yield \(b = 18 – ac\), and the third to get \(a = 6 – bc\); substituting these into the first, you get \((6 – bc) + (18 – ac) = 12\). This simplifies to \(ac + bc = 12\); but factoring out \(c\) and replacing \(a + b\) with its known value, we get \(c(a + b) = 12\), and then \(12c = 12\), so that \(c = 1\). This is one of the choices, so it looks good. But this method is entirely ad-hoc, tailored to the details of this particular problem.

The more typical approach I suggested (eliminating one variable at a time) starts by solving the first equation to get \(b = 12 – a\); then the other two equations become \(ac + 12 – a = 18\) and \(c(12 – a) + a = 6\), which simplify, respectively, to \(ac – a = 6\) and \(ac – a = 12c – 6\). Setting the two right sides equal, we get \(12c = 12\), so again \(c = 1\).

A few hours later, Doctor Rick, who had also looked at the problem, added a comment:

Jamal,

I hope you will discuss this problem further with us. There is something odd about it; perhaps you are asking because you noticed this oddity. It’s the kind of problem for which, if we do just enough work to find an answer to the question, we miss what is actually going on. I wonder whether the creator of the problem missed this as well.

I knew what he had in mind: when a problem only asks for one variable (or for one expression), rather than for a complete solution, we have seen cases where it turns out that this “solution” is only hypothetical: *if* there were a solution, then … . A simple example is one where the student may not have learned about complex numbers yet, but the equations only have non-real solutions (for which what they are asked for is valid, but they wouldn’t understand why!). But sometimes there is not a solution at all, as turns out to be the case here.

So I went further than the question asked for, and tried solving for *a* and *b* as well as *c*. Then I added my own comment:

Doctor Rick’s observation is something I missed; I had only done enough to be able to give a hint. As he says, if you follow through and try solving for a and b, rather than only for c, you will find that in fact there is

no solution, so the question is subtly wrong. (Just replace c with its value, in each equation, and the problem is obvious.)

What happens is this: Knowing that \(c = 1\), the second equation yields \(a + b = 18\), while the second becomes \(a + b = 6\). These (and also the first equation) can’t all be true! So there is actually no solution.

Jamal wrote back soon after that, saying that he had tried my first method:

Sir,

The way as Dr Peterson told I did, but as you said there is no proper solution.

What I did was

b = 18 – ac

a = 6 – bc

6 – bc + 18 – ac = 12

6 – bc + 18 – ac – 12 = 12 – 12

6 – bc + 6 – ac = 0

12 – bc – ac = 0

-c(b + a) = 0 – 12

c = 12/(b + a)seems wrong question.

Thanks for the reply.

Regards,

Jamal.

This is a slightly longer route than I took, and he hasn’t quite reached the conclusion, so I finished for him:

You didn’t quite take my suggestion to the end. You got

c=12/(b+a)

but you didn’t then replace a+b with its value, 12. The result is c = 1, right? That is apparently the expected answer.

But now take it all the way: Put that in place of c in the second and third equations, and see what happens!

So the author of the problem must not have completely solved it, but was expecting students to solve for *c* only and never notice that it is not a solution. If the problem had included “no solution”, that would have been the answer.

Now, what can we learn from this?

First, textbooks or teachers can make mistakes. You probably knew this, but perhaps hoped it wasn’t so.

Second, we need to make our assumptions explicit. When you solve an equation (or, in this case, a system of equations), you are often in a position where you are really saying, “*If* there is a solution, then it is …”. This is where **extraneous solutions** come from: They are solutions that might (in some alternate universe — that is, a somewhat different problem) have been solutions, but aren’t.

## Tricky linear systems

I discussed a somewhat similar situation in the following page, from 2017, involving a system of linear equations:

Inconsistent and Dependent? Many textbooks and online videos about a system of three equations in three variables explain thatwhen an identity is found during the elimination process, the system is dependent AND there are infinitely many solutions. I understand (or at least think I do) the definition of a dependent system for three equations and three variables, but what I am having difficulty understanding is why it can further be concluded that this system also has infinitely many solutions. All examples that I have seen both online and in textbooks show that when a 3x3 system results in an identity, it has infinitely many solutions. I confirmed this using my graphing app: each time I graphed these types of 3x3 systems, they were always three planes that intersect at a line. Butwhat if-- and I hope this makes sense -- my 3x3 system graphically showed thattwo of the equations' planes coincided, and the third equation's plane were parallel to them? For example, say I have a system like the following (I created it expressly for the purpose of my question): 2x + 4y + z = 3 (1) 4x + 8y + 2z = 6 (2) 6x + 12y + 3z = 21 (3) Eliminating x using equations (1) and (2) would result in the identity 0 = 0. With other dependent 3x3 systems, this would imply infinitely many solutions. However, eliminating x using equations (1) and (3) would result in the contradiction 0 = 12, which would imply there is no solution. This answer makes sense knowing that the planes described by (1) and (2) coincide and (3) is parallel to them. ...

What Kristi is saying is that some teaching about systems of equations ignores some important subtleties, giving the impression that you can determine that an equation has infinitely many solutions a little too easily. In my answer below I will fill in some details you may not have been familiar with. I first commented on her first paragraph:

That CAN'T be concluded without additional information. In fact, we'll see below that a system (with more than two variables)can be both dependent and inconsistent. As a result, the statement you are asking about is in fact wrong:there may NOT be infinitely many solutions.

Then I continued, focusing on her example:

You are perceptive! I use practically the same example when I teach this topic, and make similar comments about what happens if you combine different pairs of equations. (To make matters a little less obvious, I re-order the equations and throw in some zeros.) I show that changing the constant in the third equation would change the results drastically;if they never look at that equation, they will come to a wrong conclusion. ... The problem is that if you obtain an identity BEFORE YOU HAVE USED ALL THE EQUATIONS, you can't know what the other plane is doing. You have found that two planes coincide, but the third may either also coincide (the book's first case), or intersect the others in a line (its second case), OR -- and this is the case overlooked by the book, but which you thought to raise -- be parallel to the others. Two equations may be dependent, while the other is inconsistent with them.

A dependent system is one in which one equation can be obtained by combining others, so that it can be dropped from the system without changing its solution set. One way to recognize this is when you combine equations and get 0 = 0. In a system with three variables, this might mean that two of the equations are equivalent (representing the same plane), or it could mean that all points that satisfy two of the equations also satisfy the third, so that all three planes meet in a line.

What I show my students in this example is that they have to carry on with their plans to use all three equations, as if nothing odd had happened, and see what they get by combining the third equation with one of the others. If that yields an impossible equation, the system is inconsistent; otherwise, it is dependent. So to your question: what allows us to go beyond the statement that "the system is dependent" and further state that there are infinitely many solutions? The answer is:you have to go beyond finding *a* dependence, and examine the *entire system* for consistency. These are two separate issues.

So the issue here is that we need to show that a given solution satisfies **all** the equations, because the “solution” we have found may in fact not be a solution at all.

The issue in the problem we started with is just a little different, in that we did use all the *equations*, but hadn’t determined whether all the *variables* would work. So we found an “extraneous solution” only in the sense that we found a value for one variable, having stopped before finding an actual solution (ordered triple).

## A false conclusion

I discussed extraneous solutions in a February post; the first answer I discussed there, from 2013, also included a link that I left out of the post, relating to systems of equations. I gave a number of links to examples, the last being the 2006 page I’ll be looking at, with this comment:

Extraneous Routes ... This last one illustrates a category I didn't think of above, but which I happen to have discussed the other day with someone who wrote to us:Substitution in a system of equations can yield extraneous solutions.

That discussion “the other day” was later archived at

Cubic Curiosity

(which I’ll be quoting below).

The link I gave was to the following 2006 page, which is very similar. Although they aren’t primarily about systems of equations, they do use substitution.

Proof That 3 = 0? Let's say I have this: x^2 + x + 1 = 0 I can rewrite it like this: x + 1 = -x^2 and also like this: x(x + 1) + 1 = 0 So far so good, but when I use the second equation and the third I get this: x(-x^2) + 1 = 0 ---> -x^3 = -1 ---> x = 1. Now I use x = 1 in the first equation and I get 1 + 1 + 1 = 0 or 3 = 0 I must have done something incorrect but I just can't find the mistake.

Doctor Rick responded to this, first clarifying what was done there:

Let's see if I can express your reasoning more fully. I think you're saying: (1) x^2 + x + 1 = 0 (2) x + 1 = -x^2 by subtracting x^2 from each side of (1) (3) x(x+1) + 1 = 0 by factoring x from the first two terms of (1) (4) x(-x^2) + 1 = 0 by using (2) to replace (x+1) with -x^2 in (3) (5) -x^3 + 1 = 0 We see by inspection that one solution to (5) is x = 1; however, when we check this solution by substituting it in (1), we find that it is *not* a solution. Why not?

So we have several nonlinear equations that ought to be equivalent, but doing some substitutions among them yields a false result. Doctor Rick continued, with a good way to investigate anomalous results:

That's a good question! Perhaps we can learn something by finding the correct solutions to (1) and substituting them into each step of the work. Using the quadratic formula, I find that the true solutions are -1/2 + i*sqrt(3)/2 and -1/2 - i*sqrt(3)/2. You're familiar with complex numbers, right? When I said above that "*one* solution to (5) is x = 1", I had in mind that the cubic has *three* solutions; the other two are complex. In fact, those complex solutions are exactly the true solutions to (1)! Sosomehow, on the way to (5), we have introduced one extraneous solution, x = 1, in addition to the two valid solutions. This leads me to try substituting not the valid solutions, but the extraneous solution, into each step of your work: (1) x^2 + x + 1 = 0 ==> 1^2 + 1 + 1 = 0 (3 = 0, not true) (2) x + 1 = -x^2 ==> 1 + 1 = -1^2 (2 = 1, not true) (3) x(x+1) + 1 = 0 ==> 1(1+1) + 1 = 0 (3 = 0, not true) (4) x(-x^2) + 1 = 0 ==> 1(-1^2) + 1 = 0 (0 = 0, TRUE!) This tells us that the extraneous solution was added at step 4. How? Well, we can see that we replaced x+1 = 2 with -x^2 = -1, which isn't the same value, so we don't have an equivalent equation.

So, how did the substitution go wrong? Simply by forgetting that this is a chain of implications in one direction only, so that any solution of one equation is also a solution of the next one, but there is no guarantee that the new equation doesn’t also have *extra* solutions in addition. The substitution is valid when equation (1) is true; but when (1) is false, it will yield a non-equivalent equation, which may in fact be true.

## Extraneous by substitution

I explained this in the 2013 discussion I quoted in *Extraneous Routes*:

Cubic Curiosity I'm trying to find out where is the wrong step below: x^2 + x + 1 = 0 x(x + 1) + 1 = 0 [1] -x^2 = x + 1 [2] -x^3 = -1 x = 1 is not a solution for either equation [1] or [2].

Atwan starts with a quadratic equation, rearranges it in two ways, plugs one of them into the other, and gets a cubic equation whose solution is not a solution of the original. Where did this extraneous solution come from? This is identical to the 2006 question, except that he doesn’t conclude that 3 = 0.

I first restated the problem:

It's hard to follow your thinking, because you didn't explain each step you took. I'll restate this more fully, and with slight changes in the equations to make the result more dramatic. Suppose that x is a solution of x^2 + x + 1 = 0 [0] Then we can write x^2 + x = -1 Now factor this and obtain x(x + 1) = -1 [1] But starting from [0], we can also obtain x + 1 = -x^2 [2] When we substitute [2] into [1], we get x(-x^2) = -1 Simplifying, -x^3 = -1 x^3 = 1 [3] So any solution of [0] is a cube root of 1. And this is true! The solutions of [0] are the NON-REAL cube roots of 1, namely (-1 +/- i sqrt(3))/2. In fact, we can see this by factoring [3]: x^3 - 1 = 0 (x - 1)(x^2 + x + 1) = 0 Your question is, why isn't this reversible, so that ANY solution of [3], including x = 1, is a solution of [0]? And, in fact, why isn't 1 a solution of [1] or [2]?

In effect, Atwan’s work multiplied the original equation by \((x – 1)\), introducing the new solution.

The explanation lies first in a little-noticed aspect of reversibility:

The answer is that the procedure you followed is not reversible.Combining [1] and [2] to make [3] can introduce solutions of [3] that are not solutions of [1] or [2]. In particular, reversing the process would require combining, say, [2] and [3] to get [1], but 1 is not a solution of [2]. In general, when we add two equations a = b and c = d to obtain a new equation a + c = b + d, we can say that if the first two are true, so is the last. Butwe can't say that if the last is true, so are the first two. For example, when we solve a system of two linear equations by addition, we sometimes obtain the equation 0 = 0, which is true for all x and for all y. But this does not mean that each of the equations in the system is true for all x and all y; each is only true for certain ordered pairs. Thanks for an interesting question! I haven't see this particular "paradox," which is a good reminder that addition of equations has to be stated carefully to avoid the implication that it is fully reversible.

How does all this relate to our original problem? When we tried to solve it by substituting in various ways, we were creating new equations that will be true for any solution of the original system, but with no guarantee that we are not introducing new solutions in the process. The original system turns out to have no solutions, even though some of the equations derived from it do have solutions.