Solutions of $\lfloor 4x\rfloor+\lfloor 3x\rfloor=1$ 
Find all solutions of $$\lfloor 4x\rfloor+\lfloor 3x\rfloor=1$$

I have no idea as to how to go about this question. I would be grateful if somebody would please show me how to solve such questions. 
Many thanks!
 A: If $x\le0$, then the LHS is non-positive.
So $x$ has to be positive. Since both terms on the LHS are integers with $\lfloor 3x\rfloor\le \lfloor 4x\rfloor$, we have 
$$\lfloor 4x\rfloor=1\quad \text{and}\quad \lfloor 3x\rfloor =0$$
from which $\color{red}{1/4\le x\lt 1/3}$ follows.
A: Hint:
$$\lfloor4x \rfloor+\lfloor3x \rfloor=1$$
$$4x-1<\lfloor4x \rfloor\le4x$$
$$3x-1<\lfloor3x \rfloor\le3x$$
Then $$3x+4x-2<\lfloor4x \rfloor+\lfloor3x \rfloor\le3x+4x$$
$$7x-2\le1<7x$$
$$\frac 17< x\le\frac37$$
Case 1) $\frac 17< x<\frac14$
Case 2) $\frac 14\le x<\frac13$
Case 3) $\frac 13\le x<\frac37$
Answer: $$\frac 14\le x<\frac13$$
A: Let $a+b=1$ where $a,b\in \Bbb{Z}$. 
Note if $x <0$ then $\lfloor 3x\rfloor<0$ and likewise with $4x$. So $a,b\geq 0$. 
So the equation boils down to $a=1$ and $b=0$ or vice versa. Note that $\lfloor 4x\rfloor\geq \lfloor 3x\rfloor$ so you need to find $x$ so that $\lfloor 3x\rfloor=0$ and $ \lfloor 4x \rfloor = 1$.
You want $0<3x<1$ and $1\leq 4x< 2$. That is, $1/4 \leq x < 1/3$. 
