<s>[INST] The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$ The set of real values of $x$ satisfying the equation $\left[\frac{3}{x}\right]+\left[\frac{4}{x}\right]=5$,(where $[]$ denotes the greatest integer function) belongs to the interval $(a,\frac{b}{c}]$,where $a,b,c\in N$ and $\frac{b}{c}$is in its lowest form.Find the value of $a+b+c+abc$.
I tried to solve this question.Let $\left[\frac{3}{x}\right]=t$,therefore $\left[\frac{4}{x}\right]=5-t$.
$t\leq \frac{3}{x}<t+1$......(i) and $5-t\leq \frac{4}{x}<6-t$.....(ii)
Add (i) and (ii),we get $5\leq\frac{3}{x}+\frac{4}{x}<7$
$5\leq\frac{7}{x}<7$
Then i solved $5\leq\frac{7}{x}$ and $\frac{7}{x}<7$
and took the intersection of the two solution sets and got $x\in (1,\frac{7}{5}]$,so my $a+b+c+abc=48$ but the answer is given to be 20.What is wrong in my method.Please help me.
 [/INST] Suppose that $x>1$, to get $5$ its needed to add two positive integers bcz $\displaystyle \forall x>1,\left(\lfloor \frac{3}{x}\rfloor < \lfloor \frac{4}{x}\rfloor \right)$. 
So its obvious that only integer that satisfy the condition $(2,3)$ or $(1,4),$ However for the
Second couple its not find Corrosponding $x,$ bcz to have $\displaystyle \lfloor \frac{3}{x}\rfloor=1$ necessary $\displaystyle x>\frac{3}{2}$
and together with the values $\displaystyle \lfloor \frac{4}{x}\rfloor = 4\Rightarrow x<1$ Gives an empty set
Analyse Situation when $\displaystyle \lfloor \frac{3}{x}\rfloor = 2$ and $\displaystyle \lfloor \frac{4}{x}\rfloor = 3$
So we get $\displaystyle 2\leq \frac{3}{x}\leq 3$ and $\displaystyle 3\leq \frac{4}{x}\leq 4$
so solution $\displaystyle x\in \left(1,\frac{4}{3}\right]$
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