Question 633611
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What you have is what some may call a comedy of errors.  You have made mistake on top of mistake on top of mistake, and yet you managed to get half of the problem correct.


You said


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2w\ +\ 2s\ =\ 58]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ 3\ +\ s]


So far, so good. But then you said,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3\ +\ s)\ +\ s\ =\ 58]


What happened to the factor of 2 in front of the s just before the equals sign?


You should have said


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(3\ +\ x)\ +\ 2s\ =\ 58]


But even if you had that right, you would have been in trouble because you apparently forgot the Distributive Property when you asserted that *[tex \LARGE 2(3\ +\ s)\ =\ 6\ +\ s].  Nope.  *[tex \LARGE 2(3\ +\ s)\ =\ 6\ +\ 2s].  So, instead of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ +\ s\ +\ s\ =\ 58]


you should have said:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ +\ 2s\ +\ 2s\ =\ 58]


Then you made the error that got you back on track. You had *[tex \LARGE 2s\ =\ 52] and instead of ending up with *[tex \LARGE s\ =\ 26], you somehow mysteriously reinserted the missing factor of 2 to get *[tex \LARGE s\ =\ 13], which is what you would have had anyway if you had continued from:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6\ +\ 2s\ +\ 2s\ =\ 58]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4s\ =\ 52]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ s\ =\ 13].


But then you went of on a tangent again.  Why in Euler, Newton, and Leibnitz's names did you <b><i>multiply(!?)</i></b> *[tex \LARGE s] by 3 instead of adding 3 to *[tex \LARGE s] like your second original equation said?


*[tex \LARGE \ \ \ \ \ \ \ \ \ w\ =\ 3\ +\ s\ =\ 3\ +\ 13\ =\ 16]


BTW:  *[tex \LARGE 2\,\cdot\,16\ +\ 2\,\cdot\,\ 13\ =\ 32\ +\ 26\ =\ 58]


But  *[tex \LARGE 2\,\cdot\,39\ +\ 2\,\cdot\,\ 13\ =\ 78\ +\ 26\ =\ 104\ \neq\ 58]


Lesson:  ALWAYS check your work.  I do.


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