Question 627732
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In general *[tex \LARGE \frac{a\ +\ b}{c\ +\ d}\ \neq\ \frac{a}{c}\ +\ \frac{b}{d}].  And even so, how did you manage to come up with *[tex \LARGE \frac{3x}{x}\ =\ 2x\ ]?


What you need to do first is <b><i>cross-multiply</i></b>.  In general, the process is the numerator on one side of the equals sign times the denominator on the other side is equal to the denominator on the first side times the numerator on the other side, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{y}\ =\ \frac{w}{z}\ \Rightarrow\ xz\ =\ wy]


To specifics:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x\ +\ 4}{x\ -\ 2}\ =\ \frac{5}{6}]


Applying the cross-multiply process:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(3x\ +\ 4)\ =\ 5(x\ -\ 2)]


Distribute to remove parentheses:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 18x\ +\ 24\ =\ 5x\ -\ 10]


I'll let you take it from here.


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