Помогите решить задачу

$\left \{ {{x \sqrt{y}+y \sqrt{x}=6} \atop {x^{2}y+xy^{2}=20}} \right.$

Замена:

0" alt="x \sqrt{y}=t>0" align="absmiddle" class="latex-formula">

0" alt="y \sqrt{x}=m>0" align="absmiddle" class="latex-formula">

$\left \{ {{t+m=6} \atop {t^{2}+m^{2}=20}} \right.$

$\left \{ {{t^{2}+m^{2}+2tm=36} \atop {t^{2}+m^{2}=20}} \right.$

$\left \{ {{20+2tm=36} \atop {t^{2}+m^{2}=20}} \right.$

$\left \{ {{tm=8} \atop {t^{2}+m^{2}=20}} \right.$

$t^{2}+ \frac{64}{t^{2}} =20$
$\frac{t^{4}-20t^{2}+64}{t^{2}} =0$
$t^{4}-20t^{2}+64=0$

Замена: $t^{2}=k\ \textgreater \ 0$
$k^{2}-20k+64=0, D=400-4*64=144=12^{2}$
$k_{1}= \frac{20-12}{2}=4$
$k_{2}= \frac{20+12}{2}=16$

Вернемся к замене:
1) $t^{2}=4$
$t_{1}=2$
$m_{1}=4$
$t_{2}=-2<0$ - посторонний корень
2) $t^{2}=16$
$t_{3}=4$
$m_{3}=2$
$t_{4}=-4<0$ - посторонний корень

Вернемся к замене:
1) $\left \{ {{x \sqrt{y}=4} \atop {y \sqrt{x}=2}} \right.$
$\sqrt{y}= \frac{2}{x}$
$y= \frac{4}{x^{2}}$
$\frac{4 \sqrt{x} }{x^{2}}=4$
$x=1$
$y=4$
(1; 4)
2) $\left \{ {{x \sqrt{y}=2} \atop {y \sqrt{x}=4}} \right.$
$\sqrt{y}= \frac{4}{x}$
$y= \frac{16}{x^{2}}$
$\frac{16 \sqrt{x} }{x^{2}}=2$
$\frac{8 \sqrt{x} }{x^{2}}=1$
$8 \sqrt{x}=x^{2}$
$64x-x^{4}=0$
$x*(64-x^{3})=0$
$x=4$
$y=1$
(4; 1)

Ответ: (1; 4), (4; 1)