<s>[INST] no. of positive integral solutions of ||x - 1| - 2| + x = 3 What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?
My effort

||x - 1| - 2| = 3 - x
|x - 1| - 2 = 3 - x   OR   |x - 1| - 2 = x - 3
|x - 1| = 5 - x OR |x - 1| = x - 1
x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0
2x = 6 OR x $\geq$ 1
Therefore, x $\geq$ 1

But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong?
 [/INST] You should start with expanding the innermost absolute value:
$$
    \vert \vert x-1 \vert -2 \vert = \begin{cases} \vert x-3 \vert & x \geqslant 1 \\ \vert -1-x \vert & x < 1 \end{cases} = \begin{cases} x-3 & x \geqslant 1, x > 3 \\ 3-x & x \geqslant 1, x \leqslant  3 \\  1+x  & x < 1, x \geqslant -1 \\ -1-x & x < 1, x < -1  \end{cases} = \begin{cases} x-3 & x > 3 \\ 3-x & 1 \leqslant x \leqslant 3\\  1+x  & -1 \leqslant x < 1 \\ -1-x & x < -1  \end{cases}
$$
You now seek solutions for each of these branches being equal to $3-x$ within the appropriate domain, giving you three integral solutions $x=1, x=2, x=3$.
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