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 July 12th, 2012, 10:45 AM #1 Newbie   Joined: Jul 2012 Posts: 3 Thanks: 0 System of equations can anybody help 4x-4y-xy=0 2x^2+2y^2-5xy=0
 July 12th, 2012, 11:02 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: System of equations We are given the non-linear system: (1) $4x-4y-xy=0$ (2) $2x^2+2y^2-5xy=0$ From (2) we may obtain: $2x^2+2y^2-4xy=xy$ $2(x-y)^2=xy$ From (1) we obtain: $4(x-y)=xy$ Subtracting the latter result from the former, we find: $2(x-y)^2-4(x-y)=0$ $2(x-y)$$x-y-2$$=0$ So, we have 2 cases to consider: i) $x-y=0\:\therefore\=y" /> This results in $(x,y)=(0,0)$ ii) $x-y-2=0\:\therefore\=y+2" /> Substituting for x into (1), we find: $4(y+2)-4y-(y+2)y=0$ $4y+8-4y-y^2-2y=0$ $y^2+2y-8=0$ $(y+4)(y-2)=0$ This gives the solution sets: $(x,y)=(-2,-4)$ $(x,y)=(4,2)$

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