<s>[INST] Find range of $x$ satisfying $\left \lfloor \frac{3}{x} \right \rfloor+\left \lfloor \frac{4}{x} \right \rfloor=5$ Find range of $x$ satisfying $$\left \lfloor \frac{3}{x} \right \rfloor +\left \lfloor \frac{4}{x} \right \rfloor=5$$ Where $\lfloor\cdot\rfloor$ is the floor function
My try:
As far as domain of LHS is concerned we have $x \ne 0$ and since RHS is positive, we have $x \gt 0$
Now since LHS is sum of two positive integers, let us suppose:
$$\left \lfloor \frac{3}{x} \right \rfloor=m$$ and
$$\left \lfloor \frac{4}{x} \right \rfloor=5-m$$
Thus we have:
$$ m \le \frac{3}{x} \lt m+1$$
$$5-m \le \frac{4}{x} \lt 6-m$$
Adding both we get:
$$5 \le \frac{7}{x} \lt 7$$
$\implies$
$$1 \lt x \le \frac{7}{5}$$
Hence $$x \in (1, 1.4]$$
But answer in book is given as $$x \in (1,\frac{4}{3})$$
What went wrong? 
 [/INST] Your method is fine, but you dropped the inequalities prematurely. Indeed, by comparing the extreme members, we must fulfill
$$\max\left(\dfrac m3,\dfrac{5-m}4\right)<\min\left(\dfrac{m+1}3,\dfrac{6-m}4\right)$$
which is only possible with $m=2$.
Then
$$\max\left(\dfrac 23,\frac34\right)\le\frac1x<\min\left(1,1\right)$$ constrains $x$ to be smaller than $\dfrac43$.
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