<s>[INST] 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?

 [/INST] $$7t + \lfloor 2t \rfloor = 52$$
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 $t = \dfrac{369}{63} = \dfrac{41}{7} = 5 \dfrac 67$
</s>