Question

Solve the following systems of equations.
$x\sqrt{y}+y\sqrt{x}=6,x^{2}y+y^{2}x=20$

Answer 1

$x\sqrt{y}+y\sqrt{x}=6$ $x^{2}y+y^{2}x=20$

Squaring both sides

$x^{2}y+y^{2}x+2xy\sqrt{xy}=36$

$20+2(xy{)}^{3/2}=36$

$(xy{)}^{3/2}=8$

$xy=4$

$(x\sqrt{y}+y\sqrt{x}{)}^2=(x\sqrt{y}-y\sqrt{x}{)}^2+4(xy{)}^{3/2}$

$(6{)}^2-4.8=(x\sqrt{y}-y\sqrt{x}{)}^2$

$4=(x\sqrt{y}-y\sqrt{x}{)}^2$

$x\sqrt{y}-y\sqrt{x}=\pm 2$

$x\sqrt{y}+y\sqrt{x}=6$

_____________

$2x\sqrt{y}=8$ or $y$

$x\sqrt{y}=4$ or $2$

$x^{2}y=16$ or $y$

$xy=4$

So, $x=4,x=1$

$y=1,y=4$