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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+3xy-8x^2 equals 20 x^2-y^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 $\displaystyle x^2+3xy-8y^2 = 20 \\ x^2-y^2 = 15$ 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
 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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