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June 5th, 2014, 07:08 AM  #1 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0  help me solve this system of equations
Here it is: x^2+3xy8x^2 equals 20 x^2y^2 equals 15 
June 5th, 2014, 07:59 AM  #2 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
I assume it is $\displaystyle x^2+3xy8y^2 = 20 \\ x^2y^2 = 15$ Solution is intersection of two hyperbolas. $\displaystyle 8y^2 = 8x^2  120 \Rightarrow x^2 + 3xy  8x^2 + 120 = 20 \Rightarrow y = \frac{7x^2  100}{3x} \Rightarrow x^2  ( \frac{7x^2  100}{3x} )^2 = 15 \Rightarrow \\ \\ 9x^4  49x^4 + 1400x^2  10000 = 135x^2 \Rightarrow 40x^4 +1265x^2  10000 = 0 \Rightarrow x_{1,2} = \pm 4, \; \& \; x_{3,4} = \pm 2.5 \sqrt{2.5}$ 
June 5th, 2014, 09:05 PM  #3 
Newbie Joined: Jun 2014 From: rhsfhb Posts: 24 Thanks: 0 
Thanks a lot. I didnt noticed its actually that easy, thanks again for explanation 

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