1. ## two systems

Solve each of the following two systems of equations

$\displaystyle (a) x + xy + y = 2+3\sqrt2,$
$\displaystyle x^2+y^2 = 6$

$\displaystyle (b) x^2 + y^2 +2xy/(x + y)=1,$

$\displaystyle \sqrt{x+y} =x^2-y$

2. Hello,
Originally Posted by perash
Solve each of the following two systems of equations

$\displaystyle (a) x + xy + y = 2+3\sqrt2,$
$\displaystyle x^2+y^2 = 6$
multiply the first one by 2 :
$\displaystyle 2xy=4+6 \sqrt{2}-2(x+y)$

$\displaystyle \underbrace{x^2+y^2+2xy}_{(x+y)^2}=10+6 \sqrt{2}-2(x+y)$

$\displaystyle (x+y)^2+2(x+y)=10+6 \sqrt{2}$

Complete the square :
$\displaystyle (x+y+1)^2=11+6 \sqrt{2}=(3+\sqrt{2})^2$

Hence $\displaystyle x+y+1= \pm (3+\sqrt{2})$

$\displaystyle x+y=\left\{\begin{array}{ll} -4-\sqrt{2} \quad (1) \\ 2+\sqrt{2} \quad (2) \end{array} \right.$

Let's do the first case : (1)
Substitute in the equation $\displaystyle x+xy+y=2+3 \sqrt{2}$ :
$\displaystyle xy=6+4 \sqrt{2}$
So you have the system :
$\displaystyle \left\{\begin{array}{ll} xy=6+4 \sqrt{2} \\ x+y=-4-\sqrt{2} \end{array} \right.$

For the second case : (2)
Substitute $\displaystyle x+y=2+\sqrt{2}$ in the very first equation :
$\displaystyle xy=2 \sqrt{2}$
So you have the system :
$\displaystyle \left\{\begin{array}{ll} xy=2 \sqrt{2} \\ x+y=2+\sqrt{2} \end{array} \right.$

Also, be careful of the extraneous solutions.

Yah I know it's very long... I may have gone the wrong way ><
The first system gives awful solutions, but the second system gives nice solutions : (2,sqrt(2)) and (sqrt(2),2)