Question

Resolva o Sistema por cálculo de artificio

{1/x + 1/y = 3/2
{1/x² + 1/y² = 5/4

Answer 1

$\left\{\! \begin{array}{lcl} \dfrac{1}{x}+\dfrac{1}{y}&\!=\!&\dfrac{3}{2}\\\\ \dfrac{1}{x^2}+\dfrac{1}{y^2}&\!=\!&\dfrac{5}{4} \end{array} \right.$


Podemos reescrever o sistema assim:

$\left\{\! \begin{array}{ccccl} \dfrac{1}{x}&\!\!+\!\!&\dfrac{1}{y}&\!=\!&\dfrac{3}{2}\\\\ \left(\dfrac{1}{x}\right)^{\!\!2}&\!\!+\!\!&\left(\dfrac{1}{y}\right)^{\!\!2}&\!=\!&\dfrac{5}{4} \end{array} \right.$


Façamos a seguinte mudança de variáveis:

$\dfrac{1}{x}=v~~~~(v\ne 0)\\\\\\ \dfrac{1}{y}=w~~~~(w\ne 0)$


Então o sistema fica

$\left\{\! \begin{array}{ccccl} v&\!\!+\!\!&w&\!=\!&\dfrac{3}{2}\\\\ v^2&\!\!+\!\!&w^2&\!=\!&\dfrac{5}{4} \end{array} \right.$


Isole $w$ na 1ª equação:

$w=\dfrac{3}{2}-v$


Substituindo na 2ª equação, temos

$v^2+\left(\dfrac{3}{2}-v \right )^{\!\!2}=\dfrac{5}{4}\\\\\\ v^2+\left(\dfrac{3}{2}\right )^{\!\!2}-\diagup\!\!\!\! 2\cdot \dfrac{3}{\diagup\!\!\!\! 2}\cdot v+v^2=\dfrac{5}{4}\\\\\\ v^2+\dfrac{9}{4}-3v+v^2=\dfrac{5}{4}\\\\\\ 2v^2-3v+\dfrac{9}{4}-\dfrac{5}{4}=0\\\\\\ 2v^2-3v+\dfrac{4}{4}=0\\\\\\ 2v^2-3v+1=0~~~\Rightarrow~~\left\{ \!\begin{array}{l}a=2\\b=-3\\c=1 \end{array} \right.$


$\Delta=b^2-4ac\\\\ \Delta=(-3)^2-4\cdot 2\cdot 1\\\\ \Delta=9-8\\\\ \Delta=1\\\\\\ v=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ v=\dfrac{-(-3)\pm \sqrt{1}}{2\cdot 2}\\\\\\ v=\dfrac{3\pm 1}{4}$

$\begin{array}{rcl} v=\dfrac{3-1}{4}&~\text{ ou }~&v=\dfrac{3+1}{4}\\\\ v=\dfrac{2}{4}&~\text{ ou }~&v=\dfrac{4}{4}\\\\ v=\dfrac{1}{2}&~\text{ ou }~&v=1 \end{array}$


• Para $v=\dfrac{1}{2},$ encontramos

$w=\dfrac{3}{2}-\dfrac{1}{2}\\\\\\ w=\dfrac{2}{2}\\\\\\ w=1$


Mas queremos as soluções nas variáveis $x$ e $y.$ Então,

$v=\dfrac{1}{2}\\\\\\ \dfrac{1}{x}=\dfrac{1}{2}\\\\\\ \boxed{\begin{array}{c}x=2\end{array}}\\\\\\\\ w=1\\\\ \dfrac{1}{y}=1\\\\\\ \boxed{\begin{array}{c}y=1\end{array}}$


Solução 1: $(x,\,y)=(2,\,1).$


• Para $v=1,$ encontramos

$w=\dfrac{3}{2}-1\\\\\\ w=\dfrac{3}{2}-\dfrac{2}{2}\\\\\\ w=\dfrac{1}{2}$


Mas queremos as soluções nas variáveis $x$ e $y.$ Portanto, devemos ter

$v=1\\\\ \dfrac{1}{x}=1\\\\\\ \boxed{\begin{array}{c}x=1\end{array}}\\\\\\\\ w=\dfrac{1}{2}\\\\\\ \dfrac{1}{y}=\dfrac{1}{2}\\\\\\ \boxed{\begin{array}{c}y=2\end{array}}$


Solução 2: $(x,\,y)=(1,\,2).$

_________

Há dois pares ordenados que formam o conjunto solução do sistema:

$S=\big\{(2,\,1),\;(1,\,2)\big\}$


Dúvidas? Comente.


Bons estudos! :-)