Question 523977
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Your first equation says that *[tex \Large y] is equal to an expression involving *[tex \Large x].  Substitute the expression in place of *[tex \Large y] in the second equation, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x\ -\ 5y\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 15x\ -\ 5(3x\ -\ 2)\ =\ 10]


Now you have a single equation in one variable.  Solve for *[tex \Large x].  However, in this case, you do not get a definitive answer.  Solving for *[tex \Large x] results in complete elimination of the variable and the trivial result: *[tex \Large 10\ =\ 10] -- interesting, but not very.  This means that the solution set of the system has infinite elements, which is to say that the two equations have exactly the same solution set.  Graphically, they would be the same line.


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