Question 451845
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Yes, it makes no difference in the solution steps if the coefficients are fractions.  And yes, the answer is -12.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 7\ =\ \frac{1}{2}x\ +\ 1]


Add *[tex \Large -\frac{1}{2}x] to both sides.  One x minus one-half of x is one half of x.  One half of x minus one half of x is zero, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}x\ +\ 7\ =\ 1]


Add -7 to both sides.  Minus 7 plus 7 is zero, 1 minus 7 is minus 6, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}x\ =\ -6]


Multiply both sides by 2.  2 times one-half x is just x, and 2 times -6 is -12.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -12]


Done.


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