Solve for real solutions for the following system of equations:
x3+3xy2=−49x2−8xy+y2=8y−17x
My solution:
From x2−8xy+y2=8y−17x, we do some algebraic manipulation to get
(y−4x)2−16x2+x2=8y−17x(y−4x)2=8y−17x+15x2−(1)
x2−8xy+y2=8y−17x8xy=x2+y2−8y+17x−(2)
From x3+3xy2=−49, in trying to relate the y term to y−4x, we get
x3+3xy2=−49x3+3x(y−4x+4x)2=−49x3+3x[(y−4x)2+8x(y−4x)+16x2]=−49x3+3x(y−4x)2+24x2(y−4x)+48x2=−4949(x3+1)+3x[(y−4x)2+8x(y−4x)]=049(x3+1)+3x(8y−17x+15x2+8xy−32x2)=0 from (1)49(x+1)(x2−x+1)+3x[8y(x+1)−17x(x+1)]=0(x+1)[49(x2−x+1)+3x(8y−17x)]=0(x+1)[49x2−49x+49+24xy−51x2]=0(x+1)[−2x2−49x+49+3(8xy)]=0(x+1)[−2x2−49x+49+3x2+3y2−24y+51x]=0 from (2)(x+1)(x2+2x+1+3y2−24y+48)=0(x+1)[(x+1)2+3(y−4)2]=0
This is possible if and only if x=−1.
When x=−1, we get
−1−3y2=−49y2=16y=±4
