Question 190455
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{3}=2-\frac{5x-3}{x-1}]


LCD is *[tex \LARGE 3(x-1)] so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2(x-1)}{3(x-1)}=\frac{2\,\cdot\,3(x-1)}{3(x-1)}-\frac{3(5x-3)}{3(x-1)}] 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2x-2}{3(x-1)}=\frac{6x-6-15x+9}{3(x-1)}]


Now that you have two fractions with equal denominators, the numerators must be equal as well:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2x-2=6x-6-15x+9]


Collect terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x - 6x + 15x = -6 + 9 + 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 11x = 5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{5}{11}]


Checking the answer is left as an exercise for the student.  Now what were you saying about a "doozy"?


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