Rational Expression equivalent form EDIT: I know how to find the answer, but does anyone know why plugging in numbers for x does not work?
The Question: If the rational expression $\frac {3x^2}{3x-1}$ is rewritten in the equivalent form $\frac {\frac 13}{3x-1}+A$, what must expression A be in terms of x?
The four answer choices:  A) $x+ \frac 13$  B) $x+1$  C) $x-1$  D) $x-3$
The Answer:

 A) $x+ \frac 13$

I don't really know the way this question is "supposed" to be solved, so I just tried plugging in some numbers. 


*

*Plugging in 0 for x, I would get $0=\frac {-1}3+ A$

*Plugging in 1 for x, I would get $\frac 32=\frac 23 + A$.

*Plugging in 2 for x, I would get $\frac {12}{5}=\frac 53 +A$.


The equation for $x=0$ makes it seem like $A$ really is the correct answer, but plugging in 1 and 2, I wasn't sure which one, if any were correct. Pretty sure I'm just missing something really obvious here, but asking anyway. Answer Key contains the answer listed above.

You can solve the equation for A and get 
$A = \frac {3x^2-\frac 13}{3x-1}$. 
In the numerator 3 can be factored out.
$A = \frac {3(x^2-\frac 19)}{3x-1}$
$x^2-\frac 19$ is equivalent to the third binomial formula $a^2-b^2=(a-b)\cdot (a+b)$.
Therfore $ x^2-\frac 19=(x-\frac 13)\cdot (x+\frac 13)$
$A= \frac {3\cdot (x-\frac 13)\cdot (x+\frac 13)}{3x-1}=\frac { (3x-1)\cdot (x+\frac 13)}{3x-1}$

Numerical example: $x=2$
$\frac {3x^2}{3x-1}=\frac {3\cdot 2^2}{3\cdot 2-1}=\frac{12}{5}$
This result has to be the same like
$\frac{\frac13}{3x-1}+x+\frac 13=\frac{1}{9x-3}+x+\frac13$
$=\frac{1}{15}+2+\frac 13=\frac{1}{15}+\frac{30}{15}+\frac{5}{15}=\frac{36}{15}=\frac{12}{5}\quad \checkmark$
