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To solve a rational equation, say $\dfrac{3}{x-2}=\dfrac{1}{x-1}+\dfrac{7}{(x-1)(x-2)}$, the usual strategy is to multiply both sides of the equation by the least common denominator (LCD). In this case, the LCD is $(x-1)(x-2)$ and multiplying both sides of the equation yields $3(x-1)=(x-2)+7$. The question is now reduced to solving a linear equation and the solutions to this linear equation will be the solutions to the rational equation (assuming they don't create a zero in any of the denominators).

However, this method of multiplying both sides by the LCD seems funny to me. How are we 100% certain that in doing so we don't "lose" any potential solutions to the rational equation? Is there any rigorous way to prove that this method gives us $\textbf{all}$ of the solutions to a rational equation? I apologize if this question is trivial.

Intuitively, it seems as though when we multiply both sides by the LCD, we are just getting rid of the "rejected solutions" that will cause a zero in any of the denominators.

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  • $\begingroup$ You don't, which is why this method (along with raising both sides to even powers) requires one to check for extraneous solutions, something your textbook or teacher should have pointed out (assuming either applies to you). However, I believe the only way an extraneous solution can arise in this situation is when one of the solutions makes an original denominator equal to zero (because otherwise you'd be multiplying both sides by a nonzero number). $\endgroup$
    – Dave L. Renfro
    Commented Oct 31, 2020 at 17:38
  • $\begingroup$ When one of the denominators is zero it is undefined, so we should not worry about them: in a field like $\Bbb R$, multiplying both sides by a nonzero number do not change the solution set. $\endgroup$
    – player3236
    Commented Oct 31, 2020 at 17:44
  • $\begingroup$ @player3236 So I can think of it as $x \ne 2$ and $x \ne 1$ from the beginning. Thus, when we multiply both sides by the LCD, we are multiplying both sides by a non-zero number and hence not changing the solution set? $\endgroup$
    – Nicholas Roberts
    Commented Oct 31, 2020 at 17:46
  • $\begingroup$ Yes. The original solutions will remain. However if you solve the new equation and somehow get $x=1$ or $x=2$ (which is very unlikely) you can straight up reject them. $\endgroup$
    – player3236
    Commented Oct 31, 2020 at 17:48

3 Answers 3

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Losing or gaining solutions to an equation happens when you perform an operation which is either not uniquely invertible, or an inverse of such an operation. For example, with squaring we see: $$y=x\to y^2=x^2\to y=\pm x$$

The function $x^2$ is not injective and does not have a unique inverse, and so that step produces extra solutions.

However, multiplying by a given polynomial is fine at any point where the polynomial is non-zero, as the inverse of that operation is just dividing by said polynomial.

So, in your question, multiplying by $(x-1)(x-2)$ is fine, but you must discount $x=1$ and $x=2$ if they arise as solutions. In this case, they aren't, so you haven't gained any solutions.

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    $\begingroup$ I really like this response, thank you. However, do you mean that the function $x^2$ is not injective and hence when we take the inverse, we can end up with more than one element in the pre-image? You stated that its the inverse to $x^2$ that is not injective. $\endgroup$
    – Nicholas Roberts
    Commented Oct 31, 2020 at 17:54
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    $\begingroup$ Note that multiplication by $0$ does not correspond to applying an invertible function operation. That is, $f(x) = 0 \cdot x$ is not an injective function (unless the domain is a singleton or empty). See also the google search "rational equation" + "extraneous solution". $\endgroup$
    – Dave L. Renfro
    Commented Oct 31, 2020 at 17:54
  • $\begingroup$ @NicholasRoberts my mistake, good spot. You're absolutely right, since $x^2\equiv (-x)^2$. $\endgroup$
    – Rhys Hughes
    Commented Oct 31, 2020 at 17:59
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    $\begingroup$ Let $f$ be an invertible function. It may be as simple as $x = y \rightarrow f(x) = f(y) \rightarrow f^{-1}(f(x)) = f^{-1}(f(y)) \rightarrow x = y$. Hence our original solution was not lost. I see now the necessity of the operation being invertible. Thank you for all the help. $\endgroup$
    – Nicholas Roberts
    Commented Oct 31, 2020 at 18:09
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    $\begingroup$ @Nicholas Roberts: how to rigorously prove --- Regarding operations that preserve equality, see Why do we have to do the same things to both sides of an equation? and In a group $G$ with operation $\star$, can I apply $\star$ to both sides of an equation?. Regarding (essentially) operations that preserve inequality, see this 25 February 2003 sci.math post. $\endgroup$
    – Dave L. Renfro
    Commented Oct 31, 2020 at 18:33
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At worse, multiplying an equation by a common expression can introduce alien solutions (when you multiply by zero), but not lose any: if $a=b$, $ac=bc$ remains true.

Your equation has meaning only when $x\ne1$ and $x\ne2$, so if you multiply by $(x-1)(x-2)$, you will not introduce any if you keep this condition.

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  • $\begingroup$ I see. So here, $(x-1)(x-2)$ is playing the role of the non-zero $c$ you multiply both sides of the equation by. And the implication that $a=b \implies ac=bc$ for non-zero $c$ follows immediately from the field axioms. Is this a correct way to think about it? $\endgroup$
    – Nicholas Roberts
    Commented Oct 31, 2020 at 17:50
  • $\begingroup$ @NicholasRoberts: yep. Notice that $a=b\implies 0a=0b$ is also true ! No solution ever lost. $\endgroup$
    – user65203
    Commented Oct 31, 2020 at 17:50
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Declare $x\ne 1$ and $x\ne 2$ to avoid the division by zero. As you do ordinary algebraic manipulation, you get $3(x-1)-(x-2)-7 =0 \implies x=4$ which does not contradict the declarations made, this solution is correct. One can also put it in the original equation to confirm that you get $3/3=3/2$. So all is well, here in your question, you have not missed anyting.

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