Solve the simultaneous equations x^2+xy+y^2=200 x+(xy)^(1/2)+y=20

x^2 + xy + y62 =
200...............(1)

x + (xy)^1/2 + y =
20...............(2)

First we will re-wrtie equation
(2).

We will move (xy)^1/2 to the right
sides.

==> (x+ y) = 20 -
(xy)^1/2

Now we will square both
sides.

==> (x+ y)^2 = (20- (xy)^1/2)
^2

==> x^2 = 2xy + y62 = 400 - 40(xy)^1/2)
+xy

Now subtract xy from both
sides.

==> x^2 + xy + y^2 = 400 -
40(xy)^1/2)

But, from equation (1) , we know that x^2 + xy
+ y^2 = 200.

==> 200 = 400 -
40(xy)^1/2)

==> -200 =
-40(xy)^1/2

==> (xy)^1/2 = 200/40 =
5

==> xy =
25.............(3)

Now we will substitute in
(1) and (2).

==> x^2 + y^2 + 25 =
200

==> x^2 = y^2 = 175
............(4)

==> x+ 5 + y
20

==> x+ y =
15

==> y= 15-x
...............(5)

Now substitute on
(3)

==> x*y =
25

==> x( 15-x) =
25

==> 15x -x^2 =
25

==> x^2 - 15x + 25 =
0

==> x1=( 15+ 5sqrt5)/2  ==>
y1= ( 15-5sqrt5)/2

==>
x2= (15-5sqrt5) /2 ==> y2= (15+5sqrt5)/2