Question 1209168
.
I will re-formulate the problem in normal human mathematical language,
throwing out all unnecessary words


            Find all real solutions to this system of equations 
                    x + y = 10                               (1)
                    x^2 + y^2 = 62 + 2xy.          (2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
Transform second equation this way

    x^2 - 2xy + y^2 = 62,

    (x-y)^2 = 62,

    |x-y| = {{{sqrt(62)}}}.


Then the given system of equations takes the form

    x + y = 10,

    x - y = +/- {{{sqrt(62)}}}.


It falls apart in two systems of equations.  First system is

    x + y = 10,

    x - y = {{{sqrt(62)}}}.


By adding equations, you will get

    2x = 10 + {{{sqrt(62)}}}  --->  x = 5 + {{{sqrt(62)/2}}},


By subtracting equations, you will get

    2y = 10 - {{{sqrt(62)}}}  --->  y = 5 - {{{sqrt(62)/2}}}.



Second system of equations is

    x + y = 10,

    x - y = {{{-sqrt(62)}}}.


By adding equations, you will get

    2x = 10 - {{{sqrt(62)}}}  --->  x = 5 - {{{sqrt(62)/2}}},


By subtracting equations, you will get

    2y = 10 + {{{sqrt(62)}}}  --->  y = 5 + {{{sqrt(62)/2}}}.


<U>ANSWER</U>.  There are two solutions:  (x,y) = ({{{5+sqrt(62)/2}}},{{{5-sqrt(62)/2}}})

                             and   (x,y) = ({{{5-sqrt(62)/2}}},{{{5+sqrt(62)/2}}}).
</pre>

Solved.