Question 1209654
.
Find the unique pair of real numbers (x,y) satisfying
(6x^2 - 18x + 17) + (3y^2 + 6y + 11) = 28
and x + y = 20.
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        In the post by  @CPhill,  the final answer is  {(8,12)}.

        This answer  CONTRADICTS  to what is written in his post before it,  and is factually  INCORRECT.

        Also,  his post is full of non-necessary calculations.

        Below I present my solution,  which is much shorter,  simple and straightforward.

        At the end,  I will provide a correct answer.



<pre>
Starting equation is

    (6x^2 - 18x + 17) + (3y^2 + 6y + 11) = 28.


Simplify by combining like terms

    6x^2 - 18x + 3y^2 + 6y = 28 - 17 - 11,

    6x^2 - 18x + 3y^2 + 6y = 0.


Now divide both sides by 3

    2x^2 - 6x + y^2 + 2y = 0.


Substitute here y = 20-x from the second given equation

    2x^2 - 6x + (20-x)^2 + 2*(20-x) = 0,

    2x^2 - 6x + 400 - 40x + x^2 + 40 - 2x = 0,

    3x^2 - 48x + 440 = 0.


The discriminant of this quadratic equation is negative

    d = b^2 - 4ac = (-48)^2 - 4*3*440 = -2976,

which means that equation (2) has no solution.
</pre>

<U>ANSWER</U>.  &nbsp;&nbsp;There is &nbsp;{{{highlight(highlight(NO))}}} &nbsp;such pair of real numbers, &nbsp;satisfying the given conditions.


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As presented in the post, &nbsp;the problem is posed &nbsp;INCORRECTLY.


It requests to find something that does not exist in the nature,
contradicts to &nbsp;Math and can not be found.