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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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    @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.2012-09-28

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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.