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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 actuallyeliminateanswers.

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.

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

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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.

$$\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*}$$

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.

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.

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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