Question 789983
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No. 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -2\ -\ (-9)\ \not =\ 11]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -2\ -\ (-9)\ =\ -2\ +\ 9\ =\ 7]


But *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ \not =\ 11],


So what is the problem here?  Either 2 is NOT the correct replacement set, or that minus sign you have in front of the 2 in the LHS of your equation is spurious.


If the problem is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  b\ -\ (-3b\ -\ 3)\ =\ 4b\ +\ 3]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  b\ +\ 3b\ +\ 3\ =\ 4b\ + 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4b\ +\ 3\ =\ 4b\ +\ 3]


And that is true no matter what value you substitute for *[tex \Large b], so 2 would work just fine as in:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\ -\ (-3(2)\ -\ 3)\ =\ 4(2)\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2 +\ 6\ +\ 3\ =\ 8\ +\ 3]


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


But it would work just as well if *[tex \Large b\ =\ 125]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 125\ -\ (-3(125)\ -\ 3)\ =\ 4(125)\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 125 +\ 375\ +\ 3\ =\ 500\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 503\ =\ 503]



But if the problem is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -b\ -\ (-3b\ -\ 3)\ =\ 4b\ +\ 3]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -b\ +\ 3b\ +\ 3\ =\ 4b\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2b\ +\ 3\ =\ 4b\ +\ 3]


Which is only true if *[tex \Large b\ =\ 0] 


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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