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? 

The inequality after "Adding both we get:" is true, but it is not the whole story. You have lost information here, which means that not every solution to this inequality is a solution to both of the constituent inequalities.
Here is a simpler example: suppose we seek the range of solutions to
$$0<x<2$$ and $$1<x<3$$
Obviously the answer is $1<x<2$; but by your method, adding both gives
$$1<2x<5$$
which has a wider range of solutions. Your addition has lost the information that $x<2$ and $1<x$.
