Question 984332
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What makes you think that *[tex \Large x\ =\ 0] and *[tex \Large y\ =\ 0]?  *[tex \Large x] could be 100 Billion if *[tex \Large y] is negative 100 Billion from the first equation alone, and the second equation tells you that they can't both be zero because that would mean that *[tex \Large 0\ -\ 0\ +\ 2\ =\ 0] which would mean that *[tex \Large 2\ =\ 0] which would cause a catastrophic rip in the space-time continuum.  I don't think you want to be responsible for the utter destruction of the universe, do you?


Put both equations into standard form, that is:  *[tex \Large Ax\ +\ By\ =\ C]


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


Now, just since you have two of the same variable with opposite signs, you can add the two equations and eliminate the *[tex \Large y] variable altogether.  *[tex \Large x\ +\ x\ =\ 2x], *[tex \Large y\ -\ y\ =\ 0], and *[tex \Large 0\ -\ 2\ =\ -\ 2], hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2x\ -\ 0y\ =\ -2]


which is to say:


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


You should be able to handle discovering the value of *[tex \Large x] from here. Once you know *[tex \Large x] you can substitute that value into either of your original equations and then solve for *[tex \Large y].


Be careful.  The fate of the universe is in your hands.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \