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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.

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  • $\begingroup$ Break it up into cases or regions--there are three: 1) $x > 1$, 2) $0 < x < 1$, and 3) $x < 0$. You should solve each equation and then discard any found solutions that are outside of the region you are looking at. $\endgroup$ – Jared Apr 11 '15 at 6:44
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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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A shortcut formula for such sums is if $|f(x)|+|g(x)|=|f(x)+g(x)|$ then $f(x).g(x)>0$ then $[\frac{x}{x-1}][x]>=0$ which implies $x^2(x-1)>=0$.But $x^2$ is always >=0.Hence $x>1$ is the solution as well as x=0.

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  • $\begingroup$ Corrected :-)!! $\endgroup$ – user220382 Apr 11 '15 at 17:59
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Clearly, $x-1\ne0\iff x\ne1$

and $x^2=|x|(1+|x-1|)\ge0\implies x$ is real

So, $x^2=|x^2|\implies |x|(1+|x-1|-|x|)=0$

If $|x|=0, x=0$

Else $1+|x-1|-|x|=0$

We know for real $y,|y|=+y$ if $y\ge0$ and $-y$ if $y<0$

So, test for $x<0,0\le x<1,x\ge1$

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