<s>[INST] common denominator What is the common denominator of the following:
$x^2-x$ and $2x$, $x^2$
(the first is the one side, and the other 2 are another side of the formula)
The full formula is:
$1/(x^2-x) = 1/2x + 1/x^2$
P.G
That's not homework. I am a mother of 3 trying to enjoy math again
 [/INST] Your equation is
$$\frac{1}{x^2-x} = \frac{1}{2x} + \frac{1}{x^2}.$$
To find a common denominator, the first thing to look for is if any of the current denominators can be factored. The first one can, as $x^2-x = x(x-1)$, so your equation becomes
$$\frac{1}{x(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$
You then look for what each denominator is missing compared to the others. When I compare the first denominator $x(x-1)$ to the second $2x$, then the first one already has the $x$, but it is missing the factor of $2$, and when I compare the first denominator to the third $x^2$, I see that I am missing one factor of $x$ in the first to match the $x^2$ in the third. In total, I multiply the numerator and denominator of the first fraction by $2x$ to get
$$\frac{2x}{2x^2(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$
The second denominator $2x$ needs to change to become equal to the new first denominator, because this is our guess of a common denominator - but to change $2x$ to $2x^2(x-1)$, I need only multiply by $x(x-1)$, because this is what is missing. Similarly, to change the third denominator $x^2$ to $2x^2(x-1)$, I multiply by $2(x-1)$. In total, you equation becomes
$$\frac{2x}{2x^2(x-1)} = \frac{x(x-1)}{2x^2(x-1)} + \frac{2(x-1)}{2x^2(x-1)},$$
and your common denominator is $2x^2(x-1)$.
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