Question 774998
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I think you are trying to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10\ >\ \frac{2x\ -\ 5}{-4}\ >\ 5]


From now on forget you ever heard the words "Subtract", "Subtraction", "Divide", and "Division".  These operations no longer exist and you will never, for the rest of the time you spend learning and using mathematics, use them.  You always add.  If you want to do what you previously referred to as "subtraction", you add the opposite.  If you want to do what you previously referred to as "division", you multiply by the reciprocal.


The factor on the center part of the inequality is indeed *[tex \LARGE -\frac{1}{4}].  But you no longer know how to divide, right? So multiply by the reciprocal of *[tex \LARGE -\frac{1}{4}], namely *[tex \LARGE -4].  Remember to reverse the sense of the inequality because you are multiplying by a factor less than zero.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -40\ <\ 2x\ -\ 5\ <\ -20]


Add the opposite of *[tex \LARGE -5], namely *[tex \LARGE 5], to each part of the inequality.


*[tex \LARGE  \ \ \ \ \ \ \ \ \ \ -35\ <\ 2x\ <\ -15]


The factor on *[tex \LARGE x] is *[tex \LARGE 2], multiply by the reciprocal:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\frac{35}{2}\ <\ x\ <\ -\frac{15}{2}]


In interval notation:  *[tex \LARGE \left(-\frac{35}{2},\,-\frac{15}{2}\right)]


And in set builder:   *[tex \LARGE \left\{x\ \in\ \mathbb{R}\ :\ -\frac{35}{2}\ <\ x\ <\ -\frac{15}{2}\right\}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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