Question 255854
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The "other way" is substitution.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2y\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 5y\ =\ 9]


Note that in the first equation you have *[tex \Large x] by itself in the LHS.  That means that anywhere you see *[tex \Large x] in the second equation, you can replace it (hence the name "Substitution Method") with the RHS of the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\left(2y\,+\,3\right)\ -\ 5y\ =\ 9]


Distribute, collect like terms, and solve for *[tex \Large y]


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