Analyzing $\biggl\lfloor{\frac{x}{5}}\bigg\rfloor=\bigg\lfloor{\frac{x}{7}}\bigg\rfloor$ 
How many non negative integral values of $x$ satisfy the equation :$$\biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor=\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$$.

My try:
Writing few numbers and putting in the required equation.
$\underbrace{0,1,2,3,4}_{\implies 0=0},\underbrace{5,6}_{\displaystyle\times},\underbrace{7,8,9}_{\implies 1=1},\underbrace{10,11,12,13}_{\displaystyle\times} ,\underbrace{14}_{\implies 2=2},\underbrace{15,16,17,18,19,20}_{\text{LHS gives 3 but RHS gives 2}},\underbrace{21,22,23,24}_{\text{LHS gives 4 but RHS gives 3}}$.
The point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\forall x \geq 15 \implies \biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor>\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$
Hence the only solution are: $\{0,1,2,3,4,7,8,9,14\}$. Does these make sense?
I need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.
 A: Rewrite the given condition as $$5n\le x<5n+5,\quad 7n\le x<7n+7$$ where $n$ is an integer. Since $x\ge0$, $n\ge0$ as well, so the above is equivalent to $7n\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.
A: For a solution the LHS and RHS are a positive integer, let's call this $c$:
$$c = \left \lfloor \frac{x}{5} \right \rfloor = \left \lfloor \frac{x}{7} \right \rfloor$$
Then, there are integers $0 \leq a < 5, 0 \leq b < 7$, s.t.:
$$x = 5c + a = 7c + b$$
Thus:
$$a - b = 2c$$
With the restrictions to $a,b$, this is enough to find all solutions.
A: By the division rule,
$$x=5q+r=7q+s$$ with $0\le r<5,0\le s<7$.
From this we draw
$$2q=r-s\in[-6,4],$$
thus the only possibilities are with $q=0,1,2$.


*

*$q=0$: $r=s$, possible for $x\in[0,4]\cap[0,6]=[0,4]$;

*$q=1$: $5+r=7+s$, possible for $x\in[5,9]\cap[7,13]=[7,9]$;

*$q=2$: $10+r=14+s$, possible for $x\in[10,14]\cap[14,20]=[14]$.
A: Analysis:
Let us take the general form where $p>q$ :
$$
\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor
$$
Solution for above equation  exists when $\dfrac{x}{q}-\dfrac{x}{p}<1\Rightarrow x<\dfrac{p\cdot q}{p-q}$
So there is no need to check for integers greater than$\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$
Also $\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor=k$ implies
$$
\begin{align}
&x\in [k\cdot p,(k+1)p) \cap\  [k\cdot q,(k+1)\cdot q)\\
\Rightarrow\ \  &x\in [k\cdot p,(k+1)\cdot q)
\end{align}
$$
The integers which satisfy the equation are in $$[0p,1q)\ \cup\ [1p,2q)\ \cup\ [2p,3q) \cup\ ...\ \cup\ [(n-1)p,nq)$$ 
where $nq<\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$ .
