If you know that $x$ is 'just below' $7$ then, keeping your fingers crossed, just go for it!
The OP's setup allows us to write
$\tag 1 \lfloor x \rfloor = 6$
and you can begin by setting $x$ to $6$ on your 'slider bar'.
Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 6 - 1 = 41\,$)
$\tag 2 \lfloor x \times 6 \rfloor = 41$
Now keep pushing $x \lt 7$ to the right until you can write ($\, 7 \times 41 - 1 = 286\,$)
$\tag 3 \lfloor x \times 41 \rfloor = 286$
You are now left with (the rhs of the OP equation after working from the inside to the outside),
$\tag 4 x \times 286 \lt 7 * 286 = 2002$
Of course if you are saving $x = \frac{286}{41}$ from $\text{(3})$ you can push it further to the right and write
$\tag 5 x \times 286 = 2001$
so that the answer is given by
$\tag 6 x = \frac{2001}{286}$
Extra Credit: Determine if the following two equations have solutions for $x \gt 0$:
$\quad 1996=x⌊x⌊x⌊x⌋⌋⌋$
$\quad 1995=x⌊x⌊x⌊x⌋⌋⌋$
Note that we can generate similar problems.
For example, if we started by saying that $x$ is 'just below' $6$ we can crank out another corresponding 'max integer' $n$ such that
$\quad n = x⌊x⌊x⌊x⌋⌋⌋$
We would find $n$ and then ask the student to solve
$\quad 1037 = x⌊x⌊x⌊x⌋⌋⌋$
The procedure/algorithm being defined can actually be proven to produce well-defined outcomes - there is no reason to plug the 'found' $x$ back into the equation to see that 'it works'.