Is this solution correct? (A) x + y = 4 Subtracting A from B, we have x − y − x + y = 2 ( x − y ) ( x + y − 1 ) = 2 ( x − y ) ( x + y − 1 ) = ( 2 ) ( 2 )Now we have, (AA) x − y = 2 (BB) x + y − 1 = 2 This gives, x = 2 + 1 2 y = 1 2

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Is this solution correct?                    (A)                              x                +        y        =        4            Subtracting   A from   B, we have  x  −  y  −      x    +      y    =  2  (      x    −      y    )  (      x    +      y    −  1  )  =  2  (      x    −      y    )  (      x    +      y    −  1  )  =  (      2    )  (      2    )Now we have,                    (AA)                              x                −                  y                =                  2                                        (BB)                              x                +                  y                −        1        =                  2                    This gives,      x    =      2    +      1    2        y    =      1    2

Gornil2 · Accepted Answer

Do a substitution   x  =      X    2   and   y  =      Y    2  , to get 2 quadratic equations. We have the conditions that   x  ,  y  ≥  0, which also imply that   y  ≤  4  ,  x  ≤  6.Substitute one into the other and you a get   (  6  −      x    2        )    2    =  4  −  x.Solve this quartic (which doesn&#039;t seem to have nice roots), subject to   0  ≤  x  ≤  4. You will get 2 solutions. Verify if they work.

Lucian Maddox · Accepted Answer

That answer doesn&#039;t work, which you can see if you plug your values into your original equations. You made the mistake when you said      x    −      y    =      2   -- AA      x    +      y    −  1  =      2   -- BBthis doesn&#039;t follow from the previous equation because they don&#039;t both have to equal the squareroot of 2. They just have to multiply to 2. You are trying to solve 2 variable with one equation at this point. You lose information from the previous equations when you do this, so your final answer didn&#039;t work. What you might want to try instead is a substitution method, although this ends up with some very messy equations.edit: so far i end up with  0  =      y    2    −  8  y  +      y    +  10

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