please solve this two equation

Question

x^2+y=7

x+y^2=11

Answer 1

• Substitute y = 7 - x² derived from the first equation into the second equation : x + ( 7 - x² )²   = 11
• ⁴  = 11  <=====> x⁴ - 14x² + x + 38 = 0
• P(2) = 16 - 56 + 2 + 38 = 0 That means the above polynomial is divisible by (x - 2) and x = 2 is a root!
• x⁴ -14x² + x + 38 = (x - 2)(x³ + 2x² - 10x - 19) Don't bother factoring any further. ( I did try...)
• Plug x = 2 in the first equation, get 4 + y = 7 leading to y = 3 . Let's check with equation two :
• 2 + y² =  11 <=====> y = √9 = ±3 Reject y = -3 because it does not satisfy equation one.
• Final answer x = 2  &  y = 3

Answer 2

 

   X^{2 +  Y= 7}

    X + Y^{2 = 11}

 

 It is tricky: there is no standard way :

 

   Let's try This:

 

    X²  + Y = 4 +3

 

     Y² + X = 9 +2

 

     X² - 4 = Y -3

 

      Y²  - 9 = 2 - x

 

      ( X -2 ) ( X +2 ) = (Y -3)    ( 1)

 

       ( Y - 3 ) ( Y + 3 ) = - ( X -2)

 

      By substitution of ( X-2 )  in (1)

 

    - ( y-3) ( Y +3 ) = ( Y- 3 ) 

 

      ( Y - 3 ) [ (Y + 3 ) +1 ] =0   ( 2)

 

       Y - 3 = 0     Y = 3 

        Substitution in original equatio

       

        X²   +  3 = 7            X²  = 4      X = 2  ( -2 is not acceptable).

 

 

       From (2):

 

            Y + 4 = 0    Y = -4 ( is not acceptable)

 

        ( 2, 3 ) is the solution

 

             Test:   2²  + 3 = 7

 

                        3² + 2 = 11