Question 1150766
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The difference is

    p(x+y)-q(x+y) = 2x^2 + 3y^2 - 4x + 18y + 39


Continue by completing the squares

    = (2x^2 - 4x)      + (3y^2 + 18y) + 39

    = 2*(x^2 - 2x)     + 3*(y^2 + 6y) + 39 =

    = 2*(x^2 - 2x + 1) + 3*(y^2 + 6y + 9) + 39 - 2 - 27 =

    = 2*(x-1)^2        + 3*(y+3)^2        + 10.


The minimum is achieved at x= 1,  y= -3  and is equal to  10.    <U>ANSWER</U>
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Solved.