Question 261931
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I rather suspect that you mean:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{c^2\,-\,2c}\ +\ \frac{1}{2\,-\,c}\ =\ 1]


Factor the denominator in the left-most rational expression:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{c\left(c\,-\,2\right)}\ +\ \frac{1}{2\,-\,c}\ =\ 1]


Note that *[tex \Large a\ -\ b\ =\ -\left(b\ -\ a\right)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{c\left(c\,-\,2\right)}\ +\ \frac{-1}{c\,-\,2}\ =\ 1]



The LCD on the left is: *[tex \Large c\left(c\,-\,2\right)], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2\ -\ c}{c\left(c\,-\,2\right)}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ c\ =\ c\left(c\,-\,2\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c^2\ -\ c\ -\ 2\ =\ 0]


This quadratic factors:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (c\ +\ 1)(c\ -\ 2)\ =\ 0]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ -1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ c\ =\ 2]


And PLEASE, in the future, use parentheses liberally so that it is clear what it is you are trying to communicate.  It is difficult enough without having to guess at what a student means.


Your problem, given that I interpreted it correctly, should have been rendered as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ] (2/(c^2-2c)) + (1/(2-c)) = 1


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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