Question 197063
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The idea is that you never subtract, rather you add the additive inverse, and you never divide, you multiply by the reciprocal (also called the multiplicative inverse).


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x + 7 = 31 + 3x]


<b>Add</b> -7 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x + 7 + (-7) = 31 + (-7) + 3x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x = 24 + 3x]


<b>Add</b> *[tex \LARGE -3x] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x + (-3x) = 24 + 3x + (-3x)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x = 24]


<b>Multiply</b> both sides by the reciprocal of the coefficient on *[tex \LARGE x], namely *[tex \LARGE \frac{1}{12}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{1}{12}\right)\cdot 12x = \left(\frac{1}{12}\right)\cdot 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ x = 2]


The point being that Addition and Multiplication are the two basic operations, while subtraction and division are simply inverse operations defined for convenience.  Both addition and multiplication are commutative over the real numbers, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ a + b = b + a\ \forall a,\,b\ \in\ \R] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ ab = ba\ \forall a,\,b\ \in\ \R]


But subtraction and division are <i>not</i> commutative, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \exists a,\,b\ \in\ \R\ |\ a - b \neq b - a] and 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \exists a,\,b\ \in\ \R\ |\ a \div b \neq b \div a]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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