Question 808629
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There is ALWAYS a least common multiple.  (lowest common denominator is an inappropriate term in this context)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha_1x\ +\ \beta_1y\ =\ \gamma_1\ \ ]Eq. 1


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha_2x\ +\ \beta_2y\ =\ \gamma_2\ \ ]Eq. 2


Multiply Eq. 1 by *[tex \Large \alpha_2] and Eq. 2 by *[tex \Large -\alpha_1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \alpha_1\alpha_2x\ +\ \beta_1\alpha_2y\ =\ \gamma_1\alpha_2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\alpha_1\alpha_2x\ -\ \beta_2\alpha_1y\ =\  -\gamma_2\alpha_1]


Then when you add the two equations you get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0x\ +\ (\beta_1\alpha_2\ -\ \beta_2\alpha_1)y\ =\ \gamma_1\alpha_2\ -\ \gamma_2\alpha_2]


which is a single variable linear equation in *[tex \Large y].  Alternatively, you could use *[tex \Large \beta_2] and *[tex \Large -\beta_1] as your multipliers so as to end up with a single variable linear equation in *[tex \Large x], if that's what makes your socks roll up and down.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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