Solve the equation $7t+[2t] =52 $ ,where $[x]$ denotes the floor function for $x$. 
Solve the equation $7t+\left\lfloor 2t\right\rfloor =52 $.

My effort
Using the fact that for any number $x$ we have that $x=\left\lfloor x\right\rfloor+\{x\}$ (where $\{x\}$ is the fractional part of $x$) for $7t$ ,I have that:
\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor &=52 \\
\left\lfloor 7t\right\rfloor+\{7t\} +\left\lfloor 2t\right\rfloor &=52 
\end{array}
where $\{7t\}=0$ ,since we have no fractional part, and from this it also follows that $\left\lfloor 7t\right\rfloor=7t$
So the equation breaks down to 
\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor=52 \\
\left\lfloor 2t\right\rfloor =52-7t \\
\end{array}
Now, applying the definition of the floor function, I have that 
\begin{array}{c}
52-7t \le 2t <53-7t \\
52\le 9t <53 \\
52/9 \le t < 53/9 \\
\end{array}

Question
Is my effort correct? Are there other ways to approach the problem?

 A: Outline
You are absolutely right in your calculations, you only forgot to apply the other condition, that the fractional part $\{7t\} = 0$. So when you got your interval for $t$, you have to choose a $t$ satisfying this condition too.
It is easy to see that  $t = \color{blue}{\frac{41}{7}}$ is the only $t$  which will be in the interval of length $\frac{1}{9}$
Note - added explanation : $\frac{52}{9} = 5\frac{7}{9}$ and $\frac{53}{9} = 5\frac{8}{9}$. So choose $t = 5\frac{6}{7}$. 
A: The work so far shows that any solution lies in that interval, but not that every value in that interval is a solution.
On the other hand, since $\lfloor 2t \rfloor$ and $52$ are integers, if $t$ is a solution, then $7t$ must be an integer, too, that is, we can write $t = \frac{a}{7}$ for some integer $a$. Since the interval has length $\frac{1}{9}$, there is at most one such value in the interval (in fact, there turns out be exactly one), so we can solve the problem just by checking it.
A: 
Are there other ways to approach the problem?

Sure.  Let $t = m + n$, where $m$ is a multiple of $1/2$ and $0 \le n < 1/2$.  For example $\underbrace{12/5}_t = \underbrace{5/2}_m + \underbrace{1/10}_n$.  There is always 1 unique $(m,n)$ pair for each $t$.  Then you have:
$$7t + \lfloor 2t \rfloor = 52$$
$$7(m+n) + \lfloor 2(m+n) \rfloor = 52$$
$$7m+7n + \lfloor 2m+2n \rfloor = 52$$
$2m$ is an integer and $0 \le 2n < 1$ so
$$7m + 7n + 2m = 52$$
$$9m + 7n  = 52$$
The largest value $7n$ can be is $3.5$, so $52 - 3.5 < 9m \le 52$, so $5.3\bar 8 < m \le 5.\bar 7$, so $m = 5.5$.
$$49.5 + 7n = 52$$
$$n = 5/14$$
$$t = m + n = 11/5 + 5/14 = 41/7$$
A: $$7t + \lfloor 2t \rfloor = 52$$
Note that $f(t) = 7t + \lfloor 2t \rfloor$ is an increasing function. Hence there can be no more than one answer.
We note that $f(5) = 45$ and $f(6) = 54$. So the answer is somewhere between $t=5$ and $t=6$.
So lets say $t=5+\delta$ where $0 < \delta < 1$.
If $0 \lt \delta < \dfrac 12$, then
\begin{align}
   f(t) &= 52 \\
   35+7\delta + \lfloor 10 + 2\delta\rfloor &= 52 \\
   45 + 7\delta &= 52 \\
   7\delta &=7 \\
   \delta &= 1
\end{align}
So there is no solution.
If $\dfrac 12 \lt \delta < 1$, then
\begin{align}
   f(t) &= 52 \\
   35+7\delta + \lfloor 10 + 2\delta\rfloor &= 52 \\
   46 + 7\delta &= 52 \\
   7\delta &= 6 \\
   \delta &= \dfrac 67
\end{align}
So $t = 5\dfrac 67$.
CHECK:
\begin{align}
   7\left(5\dfrac 67 \right) + \left\lfloor 2 \left(5\dfrac 67 \right) \right\rfloor
   &=(35 + 6) + \left\lfloor 10 + \dfrac{12}{7} \right\rfloor \\
   &= 41 + 11 \\
   &= 52
\end{align}
As to your solution.
$$\color{red}{\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor=52 \\
\left\lfloor 2t\right\rfloor =52-7t \\
\end{array}}$$
$$\color{red}{\begin{array}{c}
52-7t \le 2t <53-7t \\
52\le 9t <53 \\
52/9 \le t < 53/9 \\
\end{array}}$$
From $\left\lfloor 2t\right\rfloor =52-7t$, we gather that $7t$ must be an integer.
So we start by making a denominator that has a factor of $7$ in it
$$\dfrac{52\cdot 7}{9\cdot 7} \le t < \dfrac{53\cdot 7}{9\cdot 7}$$
$$\dfrac{364}{9\cdot 7} \le t < \dfrac{378}{9\cdot 7}$$
Next we look for a numerator between $364$ and $377$ that is a multiple of $9$. (Then the nines will cancel and we will have a fraction with a denominator of $7$.) That number is $369$
$$\dfrac{364}{9\cdot 7} \le \dfrac{369}{9\cdot 7} < \dfrac{378}{9\cdot 7}$$
So $t = \dfrac{369}{63} = \dfrac{41}{7} = 5 \dfrac 67$
A: $\mathrm{7}{t}+\left[\mathrm{2}{t}\right]=\mathrm{52}\Rightarrow\mathrm{9}{t}\in\left[\mathrm{52},\mathrm{53}\right) \\ $
$\mathrm{2}{t}\in\left[\frac{\mathrm{104}}{\mathrm{9}},\frac{\mathrm{106}}{\mathrm{9}}\right)\Rightarrow\left[\mathrm{2}{t}\right]=\mathrm{11} \\ $
${t}=\frac{\mathrm{52}−\left[\mathrm{2}{t}\right]}{\mathrm{7}}=\frac{\mathrm{52}−\mathrm{11}}{\mathrm{7}}=\frac{\mathrm{41}}{\mathrm{7}} \\ $
