Question 654116
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You were going along sort of ok, but I don't understand why you did this:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (5)\frac{3k}{25}\ +\ 36\ =\ \frac{12k}{25}(25)]


That extra factor of 25 in the RHS does not belong there.  I'm going to start over:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3k}{5}\ +\ 44\ =\ \frac{12k}{25}\ +\ 8]


Add -44 to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3k}{5}\ =\ \frac{12k}{25}\ -\ 36]


Add *[tex \LARGE -\frac{12k}{25}] to both sides


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3k}{5}\ -\ \frac{12k}{25}\ \ =\ -36]


LCD is 25


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{5}{5}\right)\frac{3k}{5}\ -\ \frac{12k}{25}\ \ =\ -36]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{15k}{25}\ -\ \frac{12k}{25}\ \ =\ -36]


Combine like terms in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3k}{25}\ \ =\ -36]


Multiply by 25


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3k\ \ =\ -900]


Multiply by *[tex \LARGE \frac{1}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ -300]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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