Question 222904
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If (0,2) is an element of the solution set of the system:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{-3y\ +\ \,x\ =\ \ 6\cr\ \,6y\ -\ 5x\ =\ 12\right]


Then substituting 0 for *[tex \Large x] and 2 for *[tex \Large y] in either equation should result in a true statement.  So let's try it out.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3(2)\ +\ (0)\ =^?\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -6\ \neq\ 6]


So (0,2) is NOT an element of the solution set of the first equation, hence it is not an element of the solution set of the system.  Back to the drawing board.


Multiply the first equation by 5


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{-15y\ +\ 5x\ =\ 30\cr\ \,6y\ -\ 5x\ =\ 12\right]


Now add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -9y\ +\ 0x\ =\ 42]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{14}{3}]


Go back to the original two equations and multiply the first one by 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left\{-6y\ +\ 2x\ =\ \ 12\cr\ \,6y\ -\ 5x\ =\ 12\right]


Add the two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0y\ -\ 3x\ =\ 24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -8]


Hence the solution set consists of the single element: *[tex \Large \left(-8,-\frac{14}{3}\right)]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3(-\frac{14}{3})\ +\ (-8)\ =^?\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14\ -\ 8 = 6]: True


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6(-\frac{14}{3})\ -\ 5(-8)\ =^?\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -28\ +\ 40 = 12]: True


Both check, so answer checks.



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