Хелп1)

Решите задачу:

$\displaystyle\mathtt{\left\{{{x-y=5}\atop{xy=-4}}\right\left\{{{x=y+5}\atop{xy=-4}}\right~y(y+5)=-4;~y^2+5y+4=0;}\\\\\mathtt{(y+4)(y+1)=0\to\left[\begin{array}{ccc}\mathtt{x_1=y_1+5=-4+5=1}\\\mathtt{x_2=y_2+5=-1+5=4}\end{array}\right}\\\\\mathtt{OTBET:~(1;-4),~(4;-1).}$

$\displaystyle\mathtt{\left\{{{x^2+y^2=29}\atop{y^2-4x^2=9}}\right\left\{{{y^2=29-x^2}\atop{y^2=4x^2+9}}\right~29-x^2=4x^2+9;~20=5x^2;}\\\\\mathtt{x=б2~\to~y=б\sqrt{29-x^2}=б\sqrt{29-(б2)^2}=б\sqrt{29-4}=б5}\\\\\mathtt{OTBET:~(б5;б4).}$