Question 184863
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4} = 3 - \frac{2x-1}{x + 2}]


Add -3 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{4} - 3= - \frac{2x-1}{x + 2}]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{11}{4}= - \frac{2x-1}{x + 2}]


Multiply both sides by -1


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{11}{4}= \frac{2x-1}{x + 2}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 11(x + 2) =4(2x-1)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 11x + 22 =8x-4]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x = -26]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-26}{3}]


For your other one, you have two fractions set equal to each other where the denominators are equal.  You can't have two fractions with equal denominators equal to each other unless the numerators are equal also.  Just set the two numerators equal and solve for <i>x</i>.


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