System...it doesn't work!?

Question

I can't figure it out! I solve something,then I don't know what to do...
{x^2 + y^2 + xy=91
{x+y+sqrt xy=13

I solved in a way something..
For the first eq.: x^2 + y^2 +xy=91 <=>
(x+ y)^2-xy=91 <=> (x+y+sqrt xy)(x+y - sqrt xy)....

{ is a baeket..like in systems,u know!?

Answer 1

x² + y² + xy = 91
x + y + √(xy) = 13

x² + y² + 2xy = 91 + xy
(x + y)² = 91 + xy
x + y = ±√(91 + xy)

Substitute
x + y + √(xy) = 13
±√(91 + xy) + √(xy) = 13
±√(91 + xy) = 13 - √(xy)
Square both sides
91 + xy = 169 - 26√(xy) + xy
26√(xy) = 78
√(xy) = 3
xy = 9
y = 9/x

x² + y² + xy = 91
x² + (9/x)² + 9 = 91
x² +81/x² + 9 = 91
x^4 + 81 - 82x^2 = 0
(x² - 1)(x² - 81) = 0

x = ±1, ±9
y = 9/x = ±9, ±1

Check in the original equations, only the positive values work
So solution is (1,9) and (9,1)

Answer 2

x^2 + y^2 + xy=91 <-- Eq 1
x+y+sqrt xy=13 <-- Eq 2
Solve Eq 1 for y:
y^2 +xy = 91-x^2
y^2 +xy + x^2/4 = 91-x^2 +x^2/4
(y+x/2)^2 = 91-3x^2/4
y+x/2 =+/- sqrt((364-3x^2)/4)
y = -x/2 +/- .5sqrt(364-3x^2)
Substitute this in Eq 2 getting:
x+ (-x/2 +/- .5sqrt(364-3x^2)) +sqrt(x(-x/2 +/- .5sqrt(364-3x^2))) = 13
It should be a snap from here. LOL

Answer 3

There are two solutions: (x = 1, y = 9), and (x = 9, y = 1)

What does the { in front of the equations mean?

Answer 4

Square both sides of the second equation and go from there.