<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] 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
$\frac{\frac13}{3x-1}+x+\frac 13=\frac{1}{9x-3}+x+\frac13$
$=\frac{1}{15}+2+\frac 13=\frac{1}{15}+\frac{30}{15}+\frac{5}{15}=\frac{36}{15}=\frac{12}{5}\quad \checkmark$
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