<s>[INST] Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$ Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$.What will be the easiest techique to solve this sum ?
Just wanted to share a special type of equation and the fastest way to solve it.I am not asking for an answer and i have solved it in my answer given below.Thank You for viewing.
 [/INST] To me the easiest and most systematic way to solve it is to explicitly write out the absolute value which means breaking the equation into regions:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\left(\frac{1}{x - 1} + 1\right)x = \frac{x^2}{x - 1} & 1 < x < \infty \\ 
\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & 0 \leq x < 1 \\
-\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & -\infty < x < 0
\end{cases}
$$
$\frac{1}{x - 1} + 1 = \frac{1 + x - 1}{x - 1} = \frac{x}{x - 1}$ and $\frac{1}{1 - x} + 1 = \frac{1 + 1 - x}{1 - x} = \frac{2 - x}{1 - x}$ which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\frac{x^2 - x^2}{x - 1} = 0 & 1 < x < \infty \\ 
\frac{2x - x^2 - x^2}{1 - x} = 0 & 0 \leq x < 1 \\
\frac{-2x + x^2 - x^2}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
0 = 0 & 1 < x < \infty \\ 
2x\frac{1 - x}{1 - x} = 0 & 0 \leq x < 1 \\
-\frac{2x}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which means that this equation is true for all (real) values $x > 1$ and $x = 0$--there are no negative values of $x$ which satisfy this equation since the third case only has a solution at $x = 0$.
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