Question 242036
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This is uglier than a mud fence.


Multiply the first equation by *[tex \Large 2y]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{23x}{2y}\ +\ 6y\ =\ 27]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 23x\ +\ 12y^2\ =\ 54y]


Multiply the second equation by *[tex \Large 3x]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7x\ + \frac{12y}{3x}\ =\ 32]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 21x^2\ +\ 12y\ =\ 96x]


Set up to solve by substitution:


Solve the first equation for *[tex \Large x] in terms of *[tex \Large y] and *[tex \Large y^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{-12y^2\ +\ 54y}{23}]


Substitute this expression into the second equation in place of *[tex \Large x].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 21\left(\frac{-12y^2\ +\ 54y}{23}\right)^2\ +\ 12y\ =\ 96\left(\frac{-12y^2\ +\ 54y}{23}\right)]


You should be able to tell that this will simplify to a quartic in *[tex \Large y] with rational coefficients.  The rational coefficients will have large integers as numerators and denominators.


Your best bet, after you have accomplished the simplification and rearranged it into standard form, would be to plug it into a graphing calculator to determine visually whether or not you actually have any real number zeros.  Since it is a quartic, you will certainly have four roots, but the possibility exists that you will have two conjugate pairs of complex numbers.


If you are happy with approximations, your graphing calculator should handle it nicely.  But if you need the exact answers, you are going to have to struggle through the general solution to a quartic.  See the following website for a discussion of that process.


http://en.wikipedia.org/wiki/Quartic_function.  Warning:  Your response when you see the website may contain strong language and may not be suitable for young children.


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