Question 983071

You need to solve the non-linear system of two equations in two unknowns


{{{system(x^2 + y^2 - x - y = 18,
xy + 2x + 2y = 26)}}}.


Multiply the second equation by  2  (both sides)  and then add to the first one.  You will get


{{{x^2 + 2xy + y^2}}} + {{{3*(x+y)}}} = {{{18 + 2*26}}},    or


{{{(x + y)^2}}} + {{{3*(x+y)}}} - {{{70}}} = 0.


The last equation is quadratic for  (x + y)  and has the roots


a) x + y = -10     and   b) x + y = 7.


So,  combining this with the second equation of the original system, you should solve two systems


a) {{{system(x + y = -10,
xy + 2*(-10) = 26)}}}      and     b) {{{system(x + y = 7,
xy + 2*7 = 26)}}}.


or 


a) {{{system(x + y = -10,
xy = 46)}}}      and     b) {{{system(x + y = 7,
xy = 40)}}}.


The system  a)  has no real solutions. 


The system  b)  has two solutions  x=4, y=3  and  x=3, y=4. 


<B>Answer</B>. &nbsp;Two solutions are &nbsp;x=4, y=3 &nbsp;and &nbsp;x=3, y=4.