Question 316255
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I'm pretty sure you have a typo in the answer that you provided.  That leading 9 needs to be a "(" instead.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4rs\ =\ \frac{3rx\ -\ s}{2\ -\ 3x}]


Like you said, multiply by the denominator in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4rs\left(2\ -\ 3x\right)\ =\ 3rx\ -\ s]


Use the distributive property:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8rs\ -\ 12rsx\ =\ 3rx\ -\ s]


Add *[tex \LARGE -3rx] and *[tex \LARGE -8rs] to both sides and then multiply both sides by *[tex \LARGE -1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12rsx\ +\ 3rx\ =\ 8rs\ +\ s]


Factor an *[tex \LARGE x] out of the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\left(12rs\ +\ 3r\right)\ =\ 8rs\ +\ s]


Multiply both sides by *[tex \LARGE \frac{1}{12rs\ +\ 3r}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x =\frac{8rs\ +\ s}{12rs\ +\ 3r}]


Which is a perfectly good answer just like it is, but in order to match your answer, do a little factoring in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x =\frac{s\left(8r\ +\ 1\right)}{3r\left(4s\ +\ 1\right)}]


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