<s>[INST] 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.
 [/INST] $$\frac{3x^2}{3x-1} = \frac{0+3x^2}{3x-1}=\frac{(\frac{1}{3}-\frac{1}{3})+3x^2}{3x-1}=\frac{\frac{1}{3}+(-\frac{1}{3}+3x^2)}{3x-1}=\dots$$
Noting that
$$x = \frac{3x}{3}=\frac{3x+0}{3}=\frac{3x+(-1+1)}{3}=\frac{(3x-1)+1}{3}$$
Then we have
$$A = \frac{3x^2-\frac{1}{3}}{3x-1} = \frac{3\bigg(\frac{(3x-1)+1}{3}\bigg)^2-\frac{1}{3}}{3x-1}=\frac{\frac{(3x-1)^2}{3}+\frac{2(3x-1)}{3}+\frac{1}{3}-\frac{1}{3}}{3x-1}=x+\frac{1}{3}$$
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