Question
Solve the following equation : $\frac{1}{x-1}+\frac{2}{x-2}=\frac{3}{x-3}$
✓
Answer
$ \frac{1}{x-1}+\frac{2}{x-2}=\frac{3}{x-3}$
$ =\frac{1(x-2)+2(x-1)}{(x-1)(x-2)}=\frac{3}{x-3}$
$ \Rightarrow \frac{x-2+2 x-2}{x^2-2 x-x+2}=\frac{3}{x-3}$
$ \Rightarrow \frac{3 x-4}{x^2-3 x+2}=\frac{3}{x-3}$
$ \Rightarrow(x-3)(3 x-4)=\left(x^2-3 x+2\right)$
$ \Rightarrow 3 x^2-4 x-9 x+12=3 x^2-9 x+6$
$ \Rightarrow 3 x^2-13 x-3 x^2+9 x=6-12$
$ x=\frac{-6}{-4}=\frac{3}{2}=1 \frac{1}{2}$
$ =\frac{1(x-2)+2(x-1)}{(x-1)(x-2)}=\frac{3}{x-3}$
$ \Rightarrow \frac{x-2+2 x-2}{x^2-2 x-x+2}=\frac{3}{x-3}$
$ \Rightarrow \frac{3 x-4}{x^2-3 x+2}=\frac{3}{x-3}$
$ \Rightarrow(x-3)(3 x-4)=\left(x^2-3 x+2\right)$
$ \Rightarrow 3 x^2-4 x-9 x+12=3 x^2-9 x+6$
$ \Rightarrow 3 x^2-13 x-3 x^2+9 x=6-12$
$ x=\frac{-6}{-4}=\frac{3}{2}=1 \frac{1}{2}$
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