Solving $[x]=ax+1$ Given the equation $[x]=ax+1$, where $a \in \mathbb{R}$ and $[x]$ is the whole part of $x$, how do I solve for $x$?

We solve the equation in three cases:
Case (I): If $a=1$, then $[x]=x+1$. Since $[x]\in \mathbb Z$, $x+1\in \mathbb Z$ and so $x\in \mathbb Z$. Hence, $[x]=x$ and we can conclude that $x=x+1$.
Case (II): Let $a<1$. Then, since, we have $[x]\leq x<[x]+1$, $ax+1\leq x<ax+2$. It implies that $1\leq x(1-a)<2$. Therefore, $\frac{1}{1-a}\leq x<\frac{2}{1-a}$, as $a<1$.
Case (III): Let $a<1$. Then, since, we have $[x]\leq x<[x]+1$, $ax+1\leq x<ax+2$. It implies that $1\leq x(1-a)<2$. Therefore, $\frac{1}{1-a}\geq x>\frac{2}{1-a}$, as $a>1$.
