Question 1039682
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For relation A you have the ordered pairs (-8,-5) and (-3,-6)


Using the two-point form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ \left(\frac{y_1\ -\ y_2}{x_1\ -\ x_2}\right)(x\ -\ x_1) ]


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


We get



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ 5\ =\ \left(\frac{-5\ +\ 6}{-8\ +\ 3}\right)(x\ +\ 8) ]


Which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 5y\ =\ -33]


In a similar fashion, linear relation B can be expressed as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x\ -\ 5y\ =\ 43]


Solve the 2X2 system by any convenient method.  Elimination is very convenient because of the additive inverse relationship between the coefficients on *[tex \Large y]. In point of fact, a clever arithmetician can do the Elimination method just by inspection. But Cramer's Rule works just as well.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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