Question

resolva cada sistema x menos y = 6 e x ao quadrado mais y ao quadrado = 60​

Answer 1

Vamos resolver o sistema utilizando o método da substituição:

$\left\{\begin{array}{ccc}x-y&=&6\\x^2+y^2&=&60\end{array}\right\\\\\\Isolando~"y"~na~1^a~equacao:\\\\\\x-y~=~6\\\\\\\boxed{y~=~x-6}$

$Substituindo~na~2^a~equacao:\\\\\\x^2~+~(x-6)^2~=~60\\\\\\x^2~+~(x^2-12x+36)~=~60\\\\\\\boxed{2x^2-12x-24~=~0}~~~\rightarrow~Aplicando~Bhaskara\\\\\\\Delta~=~(-12)^2-4.2.(-24)~=~144+192~=~\boxed{336}$

$x'~=~\frac{12+\sqrt{336}}{2~.~2}~=~\frac{12+\sqrt{2^2~.~2^2~.~21}}{4}~=~\frac{12+4\sqrt{21}}{4}~=~\boxed{3+\sqrt{21}}\\\\\\x''~=~\frac{12-\sqrt{336}}{2~.~2}~=~\frac{12-\sqrt{2^2~.~2^2~.~21}}{4}~=~\frac{12-4\sqrt{21}}{4}~=~\boxed{3-\sqrt{21}}$

Podemos agora determinar os dois y's (y' e y'') correspondentes a x' e x'':

$\boxed{y~=~x-6}\\\\\\\\y'~=~x'-6\\\\\\y'~=~3+\sqrt{21}~-~6\\\\\\\boxed{y'~=~-3+\sqrt{21}}\\\\\\\\y''~=~x''-6\\\\\\y''~=~3-\sqrt{21}~-~6\\\\\\\boxed{y''~=~-3-\sqrt{21}}$

Temos então duas soluções para o sistema:

$\rightarrow~~(x~,~y)~=~(3+\sqrt{21}~~,~-3+\sqrt{21})\\\\\\\rightarrow~~(x~,~y)~=~(3-\sqrt{21}~~,~-3-\sqrt{21})$