SOLUTION: I need help solving this non-linear equation using the elimination method...please help. x^2+3xy+y^2=27: x^2-xy+y^2=6 any help would be greatly appreciated

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Question 603245: I need help solving this non-linear equation using the elimination method...please help. x^2+3xy+y^2=27: x^2-xy+y^2=6 any help would be greatly appreciated
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Someone has probably solved similar problems and developed an easy general procedure that I've never heard of. Maybe it would help to realize that one of the equations represents an ellipse and the other a hyperbola, and to figure out their axes of symmetry
I'll just try something and see how it works.
If you were to subtract the second equation from the first one, you would have
%28+x%5E2%2B3xy%2By%5E2%29-%28+x%5E2-xy%2By%5E2%29=27-6 --> x%5E2%2B3xy%2By%5E2-x%5E2%2Bxy-y%5E2=21 --> 4xy=21 <---> xy=21%2F4
(That's the equation of a hyperbola that has the x-axis and the y-axis as asymptotes, and tells me that x and y are both positive or both negative).
If you multiply the second equation times 3 and add the first, you would have
3%2A%28x%5E2-xy%2By%5E2%29%2B+x%5E2%2B3xy%2By%5E2=3%2A6%2B27 --> 3x%5E2-3xy%2B3y%5E2%2B+x%5E2%2B3xy%2By%5E2=18%2B27 --> 4x%5E2%2B4y%5E2=45 <---> x%5E2%2By%5E2=45%2F4
(That's the equation of a circle, centered at the origin,
with radius 3sqrt%285%29%2F2=sqrt%2845%29%2F2 ).
At this point I see two ways forward.

I can build expressions for %28x%2By%29%5E2 and %28x-y%29%5E2 , solve them for x%2By and x-y and end up with easy systems of linear equations:
%28x%2By%29%5E2=%28x%5E2%2By%5E2%29%2B2%2Axy=45%2F4%2B42%2F4=87%2F4 tells me x%2By=sqrt%2887%29%2F2 or x%2By=-sqrt%2887%29%2F2
%28x-y%29%5E2=%28x%5E2%2By%5E2%29-2%2Axy=45%2F4-42%2F4=3%2F4 tells me x-y=sqrt%283%29%2F2 or x-y=-sqrt%283%29%2F2
That seems to give too many combinations, but from the 4 sytems of linear equations
system%28x%2By=sqrt%2887%29%2F2%2C+x-y=sqrt%283%29%2F2%29, system%28x%2By=-sqrt%2887%29%2F2%2C+x-y=-sqrt%283%29%2F2%29, system%28x%2By=-sqrt%2887%29%2F2%2C+x-y=sqrt%283%29%2F2%29, and system%28x%2By=sqrt%2887%29%2F2%2C++x-y=-sqrt%283%29%2F2%29, you easily get the answers:
system%28x=%28sqrt%2887%29%2Bsqrt%283%29%29%2F4%2C+y=%28sqrt%2887%29-sqrt%283%29%29%2F4%29, system%28x=-%28sqrt%2887%29%2Bsqrt%283%29%29%2F4%2C+y=-%28sqrt%2887%29-sqrt%283%29%29%2F4%29, system%28x=-%28sqrt%2887%29-sqrt%283%29%29%2F4%2C+y=-%28sqrt%2887%29%2Bsqrt%283%29%29%2F4%29, and system%28x=%28sqrt%2887%29-sqrt%283%29%29%2F4%2C+y=%28sqrt%2887%29%2Bsqrt%283%29%29%2F4%29 .

Alternatively, I can work part of the previous option up to x%2By=sqrt%2887%29%2F2 or x%2By=-sqrt%2887%29%2F2
and then use that along with xy=21%2F4
to make two quadratic equations that will give me the answers.
We know that the quadratic equation %28z-x%29%28z-y%29=0 (z is the variable here),
with solutions x and y , can also be written as
z%5E2-%28x%2By%29z%2Bxy=0
We make one of those quadratic equations with x%2By=sqrt%2887%29%2F2 , along with xy=21%2F4 .
We get z%5E2-%28sqrt%2887%29%2F2%29z%2Bsqrt%283%29%2F3=0 , which gives us two solutions for z.
z=%28sqrt%2887%29+%2B-+sqrt%283%29%29%2F4 , both positive.
We can make x=%28sqrt%2887%29+%2B+sqrt%283%29%29%2F4 and y=%28sqrt%2887%29+-+sqrt%283%29%29%2F4 or
x=%28sqrt%2887%29+-+sqrt%283%29%29%2F4 and y=%28sqrt%2887%29+%2B+sqrt%283%29%29%2F4 .
With x%2By=-sqrt%2887%29%2F2 and xy=21%2F4 , we make
z%5E2%2B%28sqrt%2887%29%2F2%29z%2Bsqrt%283%29%2F3=0 , which gives us two negative solutions
z=%28-sqrt%2887%29+%2B-+sqrt%283%29%29%2F4 .
We can make x=%28-sqrt%2887%29%2Bsqrt%283%29%29%2F4=-%28sqrt%2887%29+-+sqrt%283%29%29%2F4 and y=%28-sqrt%2887%29-sqrt%283%29%29%2F4=-%28sqrt%2887%29%2Bsqrt%283%29%29%2F4 , or
x=%28-sqrt%2887%29-sqrt%283%29%29%2F4=-%28sqrt%2887%29+%2B+sqrt%283%29%29%2F4 and y=%28-sqrt%2887%29%2Bsqrt%283%29%29%2F4=-%28sqrt%2887%29+-+sqrt%283%29%29%2F4 .