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?