⌊3/x⌋ + ⌊4/x⌋ = 5

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Q. Find domain: $\lfloor \dfrac3x \rfloor + \lfloor \dfrac4x \rfloor = 5$

My approach:

$\dfrac3x - 1 ≤ \lfloor \dfrac3x \rfloor ≤ \dfrac3x$

$\dfrac4x - 1 ≤ \lfloor \dfrac4x \rfloor ≤ \dfrac4x$

Adding them,

$\dfrac7x - 2 ≤ \lfloor \dfrac3x \rfloor + \lfloor \dfrac4x \rfloor ≤ \dfrac7x$

$\dfrac7x - 2 ≤ 5 ≤ \dfrac7x$

$x \in (1, \frac75]$