Question 10203
 {{{x^2 - y^2 - 2x + 2y = 0 }}}
 {{{4x^2 + y^2 - 8x = 0 }}}
 In fact, the 1st equation is two-lines and the 2nd one is an ellipse
 Since {{{4x^2 + y^2 - 8x = 0}}} --> {{{4(x-1)^2 + y^2 =4 }}}
 --> {{{(x-1)^2 + y^2/4 = 1 }}} is an ellipse centered at (1,0).

 They may have at most 4 points of intersectins.

 
[The point is to cancel variable y]
 

Using complete squares to covert the 1st equation to:
 
 {{{ (x-1)^2 -(y-1)^2 = 0}}}...(1)

 Keep the 2nd equation as before  {{{4x^2 + y^2 - 8x = 0 }}} ...(2)
 From(1),we have y-1 = {{{0 +- (x-1)}}} ( +/-means postive or negative)
 When y = x-1 + 1 = x, goto (2), we have
  {{{5x^2 - 8x = 0 }}},hence x = 0 or 8/5 (why??)

 Then,we get y = 0 or 8/5 (since y = x)

 When y = -x+1 + 1 = -x+2, goto (2), we have
  {{{4x^2 + (x-2)^2 -8x = 0 }}},
 Or {{{5x^2 - 12x + 4 = 0 }}},
 Factoring: {{{(5x-2)(x-2)= 0 }}}
 So,x = 2/5 or 2 
 and then, y = 8/5 or 0.


 Thus,the four points of intersection are
 (0,0),(8/5,8/5), (2/5, 8/5), and (2,0)

 The graphs as:

 {{{ graph( 300, 300, -1, 5, -4, 4, x, -x+2, 2*sqrt(2x-x^2),
 -2*sqrt(2x-x^2) )}}} 


 Kenny