Show that $f:\mathbb{R}-\{2\}\to\mathbb{R}-\{5\}$ with $f(x)=\frac{5x-1}{x-2}$ is bijective Can anyone please help to explain the question and what actually $f: \mathbb{R} - \{2\}$  means ??
I know that bijection means  one to one function and onto both. Any idea to start up with this question?  Thanks
prove that the function $f:\mathbb{R}- \{2\} \to \mathbb{R} - \{5\}$ defined by $f(x) = (5x+1)/(x-2)$  is a bijection . (where $\mathbb{R}$ is a real number sign) 
how can i first prove that is one to one ? 
 A: In this context, $\mathbb{R}-\{x\}$ means the set of all real numbers that are different from $x$.
Injective: if $f(s)=f(t)$ then $\frac{5s+1}{s-2}=\frac{5t+1}{t-2}$ so that
$$
(5s+1)(t-2)=(5t+1)(s-2)\implies 5st-10s+t-2=5st-10t+s-2
$$
which simplifies to imply that $s=t$.
Surjective: suppose that $y$ is any number in $\mathbb{R}-\{5\}$, let us demonstrate that there is $x$ not equal $2$ such that $f(x)=y$:
$$
y=f(x)=\frac{5x+1}{x-2}=\frac{5x-10}{x-2}+\frac{11}{x-2}\iff y=5+\frac{11}{x-2}
$$
which gives
$$
x-2=\frac{11}{y-5}\iff x=2+\frac{11}{y-5}.
$$
The fraction $\frac{11}{y-5}$ is well-defined ($y\neq 5$) and nonzero, so $x$ is indeed distinct from $2$.
A: You could also show an explicit inverse, $f^{-1}: \mathbb{R} -\lbrace 5 \rbrace \rightarrow \mathbb{R} -\lbrace 2 \rbrace$, where $f^{-1}(x)= \frac{2x+1}{x-5}$ and use that 
$$f\ \  \text{is one-to-one if, and only if, has a left inverse}$$
That is, $f^{-1} \circ f(x) = f^{-1}(f(x))=x \in \mathbb{R} - \lbrace 2 \rbrace$
$$f\ \ \text{ is onto if, and only if, has a right inverse}$$
That is, $f \circ f^{-1}(x)= f(f^{-1}(x)) = x \in \mathbb{R} -\lbrace 5 \rbrace $.
