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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?

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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 Tamaroff Sep 28 '12 at 12:56
You multiplied both sides by $x-1$, which is zero if $x=1$. – Gerry Myerson Sep 28 '12 at 12:59
@50ndr33 You "multiplied by the common factor $(x-1)(x-3)$". When $x=1$, that is $0$, do you see the problem? – Pedro Tamaroff Sep 28 '12 at 13:09
@50ndr33 Just pay attention to the initial equation, write down the constraints ($x\neq 1$ in this case) and you shuld do fine. – Pedro Tamaroff Sep 28 '12 at 13:40
@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
up vote 4 down vote accepted

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.

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