Proving onto and 1-1 functions I understand the 1-1 function side of things, but I still don't really get how to prove that the function is onto
Question:
Prove that the function $f:\mathbb{R}-\{2\} \to \mathbb{R}-\{5\}$ defined by $f(x) = \dfrac{5x+1}{x-2}$ is a bijection.
So far for 1-1:
Assume that $f(a)=f(b)$, where $a,b\in \mathbb{R}-\{2\}$. Then $\dfrac{5a+1}{a-2}=\dfrac{5b+1}{b-2}$.
Cross-multiplying both sides by $(a-2)(b-2)$, we obtain $(5a+1)(b-2)=(5b+1)(a-2)$.
Simplifying, we have $5ab-10a+b-2=5ab-10b+a-2$.
Adding $-5ab$ to both sides and dividing by $-10$, we obtain $b-2=a-2$.
Adding $-2$ on both sides, we obtain $b=a$.
$\therefore f(x)=(5x+1)/(x-2)$ is 1-1.
But I'm not sure how to prove that it is onto. Can anyone help (and explain) how to prove that it is onto, so it is easy to understand?
Thanks.
 A: Let $b\in\mathbb{R}\backslash \{5\}$.  We need to find $a\in\mathbb{R}\backslash \{2\}$ so that $f(a)=b$
We need $\frac{5a+1}{a-2}=b$
$ab-2b=5a+1$
$a=\frac{2b+1}{b-5}$
You still need to confirm that $a\ne 2$.
A: A function $f : X \longrightarrow Y$ is onto or surjective if range $f = Y$, ie. for every $y \in Y$ there exists $x \in X$ such that $f(x) = y$. Does this help?
A: You should use LaTex to write maths.
Anyway, $f:X\to Y$ is said 'onto' if $Y=f(X)$, that is
$$\forall y \in Y, \exists x\in X : y=f(x)$$
That is, for every number in the codomain you can find a number in the domain which is mapped by $f$ to the first one.
Let $y$ be a real number $\not =5$. You have to find $x$ such that
$$\frac{5x+1}{x-2}=y$$
Linearize and you get $5x+1=y(x-2)$ and now solve for $x$.
$$x=\frac{-2y-1}{5-y}$$.
Since $y\not =5$, you found the number  $x$ which is mapped to $y$ for every $y$ in the codomain.
A: To show onto, you must prove that every element in your range gets used. To do this, pick $y \in \Bbb{R}-\{5\}$. We must now prove that we can always find some $x \in \Bbb{R}-\{2\}$ such that $f(x)=y$. This would mean $$\frac{5x+1}{x-2}=y \\ \implies 5x+1 = yx-2y \\ \implies 5x-yx = -2y-1 \\ \implies x(5-y) = -(2y+1) \\ \implies x = -\frac{2y+1}{5-y} \\ \implies x = \frac{2y+1}{y-5}$$ We never have to worry about dividing by zero, because $y \neq 5$. So given $y \in \Bbb{R}-\{5 \}$ we know we can find $x=\frac{2y+1}{y-5} \in \Bbb{R}-\{2 \}$. This satisfies the condition of onto.
