Question 326107
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In a word, No.  You did obtain the correct values of the coordinates of the ordered pair that is the solution set of the system.  However, you solved it by the Substitution Method, NOT the Elimination Method.


Eq. 1: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 2y\ =\ 16]


Eq. 2: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x\ +\ 7y\ =\ 1]


Multiply 1 by 7 and 2 by 2:


Eq. 3: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 28x\ -\ 14y\ =\ 112]


Eq. 4: *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ 14y\ =\ 2]


Add 3 to 4:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 38x\ +\ 0y\ =\ 114]


Here is where one of the variables (*[tex \Large y] in this case) was eliminated.


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


NOW you can substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4(3)\ -\ 2y\ =\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2y\ =\ 4]


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


See the difference?  Note that you could also have multiplied 1 by 5 and 2 by -4 and that would have allowed you to eliminate *[tex \Large x].  The result, of course, would have been the same.


By the way, the answer is most properly expressed as an ordered pair, or rather a set containing a single ordered pair, namely *[tex \Large \{\left(3,-2\right)\}]


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