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I have a simple equation: $$\frac{x}{x-3} - \frac{2}{x-1} = \frac{4}{x^2-4x+3}$$

By looking at it, one can easily see that $x \not= 1$ because that would cause $\frac{2}{x-1} $ to become $\frac{2}{0}$, which is illegal.

However, if you do some magic with it. First I factorized the last denominator to be able to simplify this: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$\frac{-(-4)\pm\sqrt{(-4)^2-4\times1\times3}}{2 \times 1}$$ $$x=1 \vee x=3$$

Then we can multiply everything with the common factor, which is $(x-1)(x-3)$ and get: $$x(x-1) - 2(x-3) - 4 = 0$$

If we multiply out these brackets, we get: $$x^2-x-2x+6-4=0$$ $$x^2-3x+2=0$$

The quadratic formula gives $x = 1 \vee x=2$. We already know that $x$ CANNOT equal to 1, but we still get it as an answer. Have I done anything wrong here, because as I see it, this is the same as saying that:

$$\frac{x}{x-3} - \frac{2}{x-1} = \frac{4}{x^2-4x+3}$$ $$=$$ $$x(x-1) - 2(x-3) - 4 = 0$$ which cannot be true, because the two doesn't have the same answers. What am I missing here?

Michael Hardy
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Friend of Kim
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    The equations are not equivalent, as you note. But if you introduce the clarification $x\neq 1$ on the second, they are. – Pedro Sep 28 '12 at 12:56
  • @PeterTamaroff Ok, but how can they not be the same if I haven't done anything illegal? If they aren't the same, one can't put an equal sign between all the steps... – Friend of Kim Sep 28 '12 at 12:58
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    You multiplied both sides by $x-1$, which is zero if $x=1$. – Gerry Myerson Sep 28 '12 at 12:59
  • @GerryMyerson Could you please clarify a little bit? – Friend of Kim Sep 28 '12 at 13:01
  • Think about a simpler example, $1/x=2/x$, which obviously has no solution. Multiply by $x$ to get $x=2x$, with the solution $x=0$. Can you see the problem here? – Gerry Myerson Sep 28 '12 at 13:02
  • Yes, you have done only thigs that are considered "legal", and you end up with an equation that is not equal to the one you started with, and you get an answer 0 that is not allowed in the first equation. So, are you saying that you cannot necessarily multiply the whole equation and get a new equation that is equal to the first one? – Friend of Kim Sep 28 '12 at 13:06
  • @GerryMyerson If you multiply by $x$ you get $2=1$. I don't think that is a good example. – Pedro Sep 28 '12 at 13:08
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    @50ndr33 You "multiplied by the common factor $(x-1)(x-3)$". When $x=1$, that is $0$, do you see the problem? – Pedro Sep 28 '12 at 13:09
  • @PeterTamaroff Yes, now I see the problem! Thanks! I'm used to the fact that you can do anything to the x-es because you don't know what their value is. – Friend of Kim Sep 28 '12 at 13:11
  • @Peter, yes, sorry, make that "multiply by $x^2$". – Gerry Myerson Sep 28 '12 at 13:16
  • @50ndr33 What on Earth did you write there? =) – Pedro Sep 28 '12 at 13:17
  • But then you can do $\frac{1}{x} = \frac{2}{x} | \times x$ and get $2=1$. – Friend of Kim Sep 28 '12 at 13:22
  • @PeterTamaroff I feel really stupid right now... I was just too caught up trying to make it become $x=2x$... – Friend of Kim Sep 28 '12 at 13:31
  • @PeterTamaroff and – Friend of Kim Sep 28 '12 at 13:37
  • @GerryMyerson But if there are "traps" like this, how does one discover them when the problems get more complex than just the simple problem I wrote? – Friend of Kim Sep 28 '12 at 13:38
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    @50ndr33 Just pay attention to the initial equation, write down the constraints ($x\neq 1$ in this case) and you shuld do fine. – Pedro Sep 28 '12 at 13:40
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    @50ndr33: Whenever you even write down $\frac a b$, make sure that $b\ne 0$. And whenever you multiply an equation by something, be sure that this something is $\ne 0$ (or you get only $\Rightarrow$ instead of $\Leftrightarrow$). – Hagen von Eitzen Sep 28 '12 at 14:01
  • Please add your comments as an answers, as it is more useful for other people that come from Google. :) – Friend of Kim Sep 28 '12 at 14:06
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    @Friend, one method is, after you have found the "solutions", plug them back into the original equation to see whether they really work, and discard as extraneous any that don't. – Gerry Myerson Sep 28 '12 at 22:30

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If $\dfrac AB = 0$ then $A=0\cdot B$. But you can't say that if $A=0\cdot B$ then $\dfrac AB=0$ unless you know that $B\ne 0$. So if $A$ and $B$ are complicated expressions that can be solved for $x$, there may be values of $x$ that make $B$ equal to $0$, and if they also make $A$ equal to $0$, then they are solutions of the equation $A=0\cdot B$, but not of the equation $\dfrac AB=0$.

"If P then Q" is not the same as "If Q then P".

Another way of putting it is that this explains why "clearing fractions" is one of the operations that can introduce "extraneous roots". Perhaps more well known is that squaring both sides of an equation can do that.

Michael Hardy
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