Question 806319
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Yes, you are wrong.  You substituted the value 4 when you were asked to use the value -4.  Two different numbers.  But then you did the arithmetic incorrectly and thereby inadvertently came up with the correct assessment that the number given is, indeed, an element of the solution set of the equation.


Here it is done correctly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ -\ 2x\ =\ 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ -\ 2(-4)\ =^?\ 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ -\ (-8)\ =^?\ 15]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ +\ 8\ =\ 15]


Since the result of the substitution is a true statement, the value is an element of the solution set.


Which god? 


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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