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June 5th, 2014, 07:08 AM   #1
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help me solve this system of equations

Here it is:
x^2+3xy-8x^2 equals 20

x^2-y^2 equals 15
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June 5th, 2014, 07:59 AM   #2
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Math Focus: Geometry
I assume it is
$\displaystyle x^2+3xy-8y^2 = 20 \\ x^2-y^2 = 15$

Solution is intersection of two hyperbolas.

$\displaystyle 8y^2 = 8x^2 - 120 \Rightarrow x^2 + 3xy - 8x^2 + 120 = 20 \Rightarrow y = \frac{7x^2 - 100}{3x} \Rightarrow x^2 - ( \frac{7x^2 - 100}{3x} )^2 = 15 \Rightarrow \\ \\ 9x^4 - 49x^4 + 1400x^2 - 10000 = 135x^2 \Rightarrow -40x^4 +1265x^2 - 10000 = 0 \Rightarrow x_{1,2} = \pm 4, \; \& \; x_{3,4} = \pm 2.5 \sqrt{2.5}$
Thanks from wannabemathlete
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June 5th, 2014, 09:05 PM   #3
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Thanks a lot. I didnt noticed its actually that easy, thanks again for explanation
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