# Why does this equation have different number of answers?

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

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 $\dfrac AB=0$.