Rational Expression Simplification The function:
$$f(x) = \frac{3x - 4}{x^2 - 2x}$$
is simplified to:
$$f(x) = \frac{2}{x} + \frac{1}{x - 2}$$
How? And in what way?
 A: We would like to write $\dfrac{3x-4}{x^2-2x} = \dfrac{3x-4}{x(x-2)}$ in the form $\dfrac{A}{x} + \dfrac{B}{x-2}$ for some constants $A,B$. 
Now, we just need to solve for $A,B$. First, let's get rid of fractions: 
$\dfrac{A}{x} + \dfrac{B}{x-2} = \dfrac{3x-4}{x(x-2)}$
$A(x-2)+Bx = 3x-4$
This equality must be true for all values of $x$, so it must be true for $x = 0$ and $x = 2$:
Plugging in $x = 0$ gives us $A \cdot (0-2) + B \cdot 0 = 3 \cdot 0 - 4$, i.e. $-2A = -4$. Hence, $A = 2$. 
Plugging in $x = 2$ gives us $A \cdot (2-2) + B \cdot 2 = 3 \cdot 2 - 4$, i.e. $2B = 2$. Hence, $B = 1$. 
Therefore, $\dfrac{3x-4}{x(x-2)} = \dfrac{2}{x} + \dfrac{1}{x-2}$.
You can also solve $A(x-2)+Bx = 3x-4$, by writing it as $(A+B)x+2A = 3x-4$, and equating coefficients. Then, you get the equations $A+B = 3$ and $2A = -4$. Solving this linear system gives $A = 2$ and $B = 1$ as you got with the first method. 
Note: This technique is called Partial Fraction Decomposition. I suggest you Google that term for more information on this method. 
A: $$f(x)=\frac {3x-4}{x^2-2x}\\
=\frac {3x-4}{x(x-2)}\\
=\frac Ax +\frac B{x-2}\\
=\frac {A(x-2)+Bx}{x(x-2)}$$
Equation numerators:
$$3x-4=A(x-2)+Bx\\
=(A+B)x-2A$$
gives $A=2,B=1$, i.e. 
$$f(x)=\frac {3x-4}{x^2-2x}=\frac 2x +\frac 1{x-2}$$
