Show that the function is a bijection and find the inverse function. Is my answer correct for the following question?
Show that the function $f : \mathbb{R} − {3} \to \mathbb{R} − {2}$ defined by $f(x) = \frac{2x−3}{x-3}$ is a bijection, and find the inverse function. I don't necessarily have to prove anything.
To show it's a bijection, I need to show that it's one-to-one and onto.
A function is one-to-one iff. $f(a)=f(b)$ implies that $a=b$
$\frac{2a-3}{a-3} =\frac{2b-3}{b-3}$ 
$(2a-3)(b-3)=(a-3)(2b-3)$
$2ab-6a-3b+9=2ab-3a-6b+9$
$-3b=-3a$
$b=a$
When $f(a)=f(b)$, $b=a$, therefore, this function is one-to-one.
A function is onto iff. for every $b\in B$, there is an $a \in A$ such that $f(a)=b$
For any $b\in R$-{2}, there's $a\in R$-{3} such that $f(a)=b$
Since $f(x)$ is one-to-one and onto, it is a bijection.
Finding the inverse:
$f^{-1}(x)=f(y)$
$f(y)=x$
$x= \frac{2y-3}{y-3}$
$xy-2y=3x-3$
$y(x-2)=3x-3$
y= $\frac{3x-3}{x-2}$
 A: There's a couple parts in your answer that are confusing:
When proving $f$ is 1-1, you have $(2a−3)(b−3)=(a−3)(2b−3)$ implies $b-3 = a-3$, but I'm not sure how you're making that jump. I would instead note that you have:
$$(2a-3)(b-3) = (a-3)(2b-3) \longrightarrow 2ab - 6a -3b + 9 = 2ab -3a -6b + 9$$
Then by canceling the $2ab$ and the 9, you will get $3b = 3a$, or $b=a$ as desired.
Then, when proving that $f$ is onto, we need to start with an output value $y$ and show that there is some $x$ in the domain such that $f(x) = y$. In your explanation, you've discussed for which $x$ values the function is well-defined; ie, the domain of the function. 
Instead, we let $y \in \mathbb{R} - \{2\}$. Then $y = \frac{2x-3}{x-3}$, so $y(x-3) = 2x - 3$, or $3 - 3y = 2x - yx = (2-y)x$. Then $x = \frac{3-3y}{2-y}$, which is well defined for all $y \in \mathbb{R}-\{2\}$. Note too that this $x$ will never be 3, or we'd have $6=3$. Thus for all $y$ in the range, we can find $x \in \mathbb{R} - \{3\}$, $x = \frac{3-3y}{2-y}$ such that $f(x) = y$, so our function is onto. 
Your inverse function looks correct to me.
A: For the injectivity part, you say $$(2a-3)(b-3)=(a-3)(2b-3),$$ and then jump to $$a-3=b-3.$$ I do not think that the first follows immediately from the second. Or if it does in some way I am not seeing, you should probably add some intermediate steps.
For surjectivity, you need to prove that for every $b\in\mathbb R-\{2\}$, there is an $a\in\mathbb R-\{3\}$ so that $f(a)=b$. What you proved (or rather, stated) is that for all $a\in\mathbb R-\{3\}$, $f(a)\in\mathbb R$. Notice the difference? Examine the definition of surjectivity again.
