<PRE><B>Question 7773</B>
 To solve:
 {{{x^(1/2)+y=7}}} ...(1)
{{{x+y^(1/2)=11}}} ...(2)

 From(1): x^(1/2) = 7 - y ,
 Squaring both sides:
  x = (7-y)^2 = y^2 - 14y + 49,
 Goto (2): y^2 - 14y + 49 + y^(1/2) = 11,
 Let w = y^1/2, so  y = w^2, 
 so we have w^4 - 14 w^2  + w + 38 = 0,
 Use syntheic division,we have
  1 + 0-14 + 1 + 38 (2
    + 2 +4 -20 -38
 ----------------------
  1 + 2 -10 -19 + 0

 Hence,(w-2)(w^3 + 2w^2-10w -19) = 0.
 Unfortunately, we cannot factor w^3 + 2w^2-10w -19,
 and the three roots are in very bad form.
 
Note,when w =2, y= 4 and x =  16 - 56 + 49 = 9
 So, here I only give you one set of solution 
 x= 9 and y = 4.

  Kenny</PRE>