Question 1209168
<br>
{{{x+y=10}}}
{{{x^2+y^2=62+2xy}}}<br>
Whenever the given equations have terms of {{{x^2}}}, {{{y^2}}}, and {{{2xy}}}, one thing you want to look at is using one of these:<br>
{{{(x+y)^2=x^2+2xy+y^2}}}
{{{(x-y)^2=x^2-2xy+y^2}}}<br>
For this problem....<br>
{{{x^2+y^2=62+2xy}}}
{{{x^2-2xy+y^2=62}}}
{{{(x-y)^2=62}}}<br>
Solve the first given equation for y in terms of x and substitute in this last equation.<br>
{{{y=10-x}}}
{{{x-y=x-(10-x)=x-10+x=2x-10}}}
{{{(x-y)^2=(2x-10)^2=62}}}
{{{4x^2-40x+100=62}}}
{{{4x^2-40x+38=0}}}<br>
That quadratic does not factor, so use the quadratic formula to find<br>
{{{x=(40+sqrt(992))/8}}} and {{{x=(40-sqrt(992))/8}}}<br>
Partially simplified, those roots are<br>
{{{5+sqrt(992)/8}}} and {{{5-sqrt(992)/8}}}<br>
The given equations are symmetrical in x and y, so the two solutions are that x and y are equal to, in either order, {{{5+sqrt(992)/8}}} and {{{5-sqrt(992)/8}}}<br>
Those numbers, confirmed by graphing the system of equations using desmos.com, are approximately 8.937 and 1.063.<br>
ANSWERS (approximately): (8.937,1.603) and (1.063,8.937)<br>