Question 261896
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I have no idea what "quardents" are, nor how to "pare" them if I did know what they were.  Give me a potato and a small knife and I can pare with the best of them, but this isn't a cooking show, is it?  Now if what you really want is an ordered <b>pair</b> of <b>coordinates</b> that represents the solution set of the system of linear equations that you presented, I <b><i>can</i></b> help you with that.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ -\ y\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 3]


Note that the coefficients on the *[tex \Large y] terms are additive inverses already (namely -1 and 1).  So all you have to do to start is add the equations, term by term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ +\ 0y\ =\ 4]


Multiply by *[tex \Large \frac{1}{2}]


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


Substitute this value for *[tex \Large x] into either of the original equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (2)\ -\ y\ =\ 1]


And then solve for *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 1]


Construct the ordered pair from the values that you have just derived:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x,\,y\right)\ =\ \left(2,\,1\right)]


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