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I'm currently working through an algebra book, and during the chapter about rational expressions and inequalities, the author has a side note in which he states:

Never divide both sides of the equation by a variable, even if you're doing it to try to solve a rational equation, because there's a very real danger that you will actually eliminate answers.

The reason I'm asking this question is because I didn't understand how answers can be eliminated when dividing both sides of an equation by a variable.

Can you show me an example of how this can happen and why?

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    $\begingroup$ Solve $x^2=x$. Divide both sides by $x$. We get $x=1$. We have lost the solution $x=0$ of the original equation. But of course we can divide, as long as we keep track of solutions we are losing. $\endgroup$ Sep 27, 2011 at 19:35
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    $\begingroup$ $x(x^2 + 2x + 1) = 0$ divide by $x$ and we get $ (x+1)^2 = 0$ or $x = -1$ but we eliminated $0$ as an answer. You have to be careful because you might divide by $0$ which is not allowed $\endgroup$
    – Deven Ware
    Sep 27, 2011 at 19:37
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    $\begingroup$ I'm currently studying calculus and I see math teachers dividing both sides by variables all the time, so I don't understand how to know which situations it's okay in. $\endgroup$ Sep 7, 2015 at 2:13
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    $\begingroup$ @Kyle For the sake of future readers of this question, here's why. (To my understanding.) It is okay to do such division, but then the solution no longer becomes valid whenever dividing by 0 occurs as a result of doing so. In calculus, you will frequently note that when such division is performed, it is always noted to the side that $x \neq 0$ for example. $\endgroup$
    – Bassinator
    Sep 29, 2016 at 0:59
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    $\begingroup$ @AndréNicolas Is valid to extend your statement to say that when we multiply by a variable, we have to keep track of which solutions we are adding? And what would that look like? $\endgroup$
    – hypehuman
    Mar 6, 2020 at 11:20

5 Answers 5

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When you divide, you are implicitly assuming that the number you are dividing by is not equal to zero. By dividing, you are excluding the possibility that the number in question is zero, and as such you may be eliminating correct answers.

For a very simple example, consider the case of the equation $x^2-x=0$.

There are two answers: $x=0$, and $x=1$. However, if you "divide by the variable", you can end up doing this: $$\begin{align*} x^2 - x & = 0\\ x^2 &= x &&\text{(adding }x\text{ to both sides)}\\ \frac{x^2}{x} &= \frac{x}{x} &&\text{(divide by }x\text{, which assumes }x\neq 0)\\ x &= 1. \end{align*}$$ So you "lost" the solution $x=0$, because when you divided by $x$, you implicitly were saying "and $x\neq 0$". In order to "recover" this solution, you would have to consider "What happens if what I divided by is equal to $0$?"

For a more extreme example, consider something like $$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=0.$$ Since a product is equal to $0$ if and only if one of the factors is equal to $0$, there are six solutions to this equation: $x=1$, $x=2$, $x=3$, $x=4$, $x=5$, and $x=6$. Divide both sides by $x-1$, and you lose the solution $x=1$; divide both sides by $x-2$, you lose $x=2$. Continue this way until you are left with $x-6=0$, and you lost five of the six solutions. And if then you go ahead and divide by $x-6$, you get $1=0$, which has no solutions at all!

Whenever you divide by something, you are asserting that something is not zero; but if setting it equal to $0$ gives a solution to the original equation, you will be excluding that solution from consideration, and so "eliminate" that answer from your final tally.

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    $\begingroup$ + for showing that dividing by anything is dangerous, not only by a variable. $\endgroup$
    – lisyarus
    Oct 21, 2016 at 16:51
  • $\begingroup$ I have a question about this: is it the fact that whenever you divide by a variable or variable expression, one of the solutions was 0 somewhere so therefore you will always divide by 0 or is it that you can potentially divide by 0? $\endgroup$
    – user29418
    Jan 27 at 1:26
  • $\begingroup$ @user29418 If you divide by so,ething, you are asserting that the thing you are dividing by is not zero. You are essentially adding a condition to the equation you are solving. This condition may not matter (it may be superfluous), $\endgroup$ Jan 27 at 4:19
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$$\begin{align*} x^2 - x & = 0\\ x^2 &= x &&\text{(adding }x\text{ to both sides)}\\ \frac{x^2}{x} &= \frac{x}{x} &&\text{(divide by }x\text{, which assumes }x\neq 0)\\ x &= 1. \end{align*}$$

This assumes $x\ne0$. Check this out:

$$\begin{align*} x^2 - x & = 0\\ x(x-1) &= 0 &&\text{(factor)}\\ x=0 \quad &\text{or} \quad (x-1)=0 && \text{(by some rule in } \mathbb{R} \text{ that says } ab=0 \implies a=0 \lor b=0 \text{)}\\ x=0 &\lor \ x=1.\\ \end{align*}$$

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If you do it correctly, you most certainly can divide both sides of an equation by a variable.

As long as $x \ne 0$, $\dfrac 1x$ exists; and, as long as $x \ne 0,\; xA = xB \iff A = B$.

Here is the logic you need to go through to divide both sides by $x$ correctly.

(1) We want to find all solutions to $x^2 = 3x$.

(2) Clearly $x = 0$ is a solution.

(3) Suppose $x \ne 0$. Then we can divide both sides by $x$.

(4) We get $x=3$.

(5) So the solution set is $x \in \{0, 3\}$.

Note that, dividing both sides by $x$ without accounting for the case $x=0$ means that you are "throwing away" that particular solution.

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Another good example is shown when solving a trigonometric equation.

There are four solutions to the equation: $\sin x \tan x = \sin x$,

\begin{align} \sin x \tan x &= \sin x \\ \sin x \tan x - \sin x &= 0 \\ \sin x (\tan x - 1) &= 0 \end{align}

$\sin x = 0$ or $\tan x = 1$. The solution set is $\{0^\circ, 45^\circ, 180^\circ, 225^\circ\}$.

However, trying to solve the equation by dividing each side by $\sin x$ would lead to just $\tan x = 1$, which would only give you $x = 45^\circ$ or $x = 225^\circ$. The other two solutions would not appear. The missing solutions are the ones that make the divisor ($\sin x$) equal zero. For this reason we avoid dividing by a variable expression.

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Assuming division is not the correct step, none of the answers clearly mention what to do instead. Suppose you end up with $$ f(x)g(x)=f(x)h(x),$$ where $f(x)$ is the common factor. Instead of dividing by $f(x)$, you rewrite the equality as: $$ f(x)(g(x)-h(x)) = 0.$$ Now $f(x)=0 \vee g(x)-h(x)=0$.

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  • $\begingroup$ I don't think your first sentence is fair. All of the answers above are giving examples of your solution. $\endgroup$
    – M Turgeon
    Oct 21, 2016 at 16:55

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