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Find the solution of $$1 \le \left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor \lt 2$$ My try:

The only thing i know is that $$\left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor \in \Bbb Z$$ so $$\left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor = 1$$

but i don't know how to continue.

Any hints?

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Hint: $1 \leq \lfloor x \rfloor <2$ iff $1 \leq x <2$ so you have to solve $1-x \leq |\frac {1+x} {x-3}| < 2-x$. Consider the cases $\frac {1+x} {x-3}>0$ and $\frac {1+x} {x-3} \leq 0$ separately.

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