Question 650342
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If you have a rational expression of the form *[tex \LARGE \frac{A}{B}\ \leq\ 0] then *[tex \LARGE A] and *[tex \LARGE B] must be of opposite signs.  If they were the same sign, then you would have a positive value for the rational expression which would make the relation false.


Therefore we either have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3y\ +\ 11\ \leq\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 8\ >\ 0]


Note that we must not allow the denominator to equal zero.


Solving these two inequalities gives:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \leq\ -\frac{11}{3}]


AND


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ >\ 8]


Which is absurd -- *[tex \LARGE y] cannot simultaneously be larger than 8 and smaller than a number in the vicinity of -4.


Try it the other way:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3y\ +\ 11\ \geq\ 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 8\ <\ 0]


Again, we must not allow the denominator to equal zero.


Solving these two inequalities gives:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ \geq\ -\frac{11}{3}]


AND


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ <\ 8]


Which gives the solution set: *[tex \LARGE \left\{y\,|\,y\,\in\,\mathbb{R},\,-\frac{11}{3}\,\leq\,y\,<\,8\right\}]



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