Question

Me ajudeeeemmm, sei q tá borrada, se não entenderem pergunte nos coment, pra hj, por favorzinhooo
X+4y=7
X ao quadrado+ Y ao quadrado= 10
(Isso é um sistema)

Answer 1

Resolver o sistema:

$\left\{ \!\begin{array}{lc} x+4y=7&~~~~\mathbf{(i)}\\\\ x^2+y^2=10&~~~~\mathbf{(ii)} \end{array} \right.$


Isole x na equação (i) e substitua na equação (ii):

$x=7-4y\\\\\\ (7-4y)^2+y^2=10$


Desenvolva o quadrado da diferença via produtos notáveis:

$7^2-2\cdot 7\cdot 4y+(4y)^2+y^2=10\\\\ 49-56y+16y^2+y^2=10\\\\ 49-56y+17y^2-10=0\\\\ 17y^2-56y+39=0~~~\Rightarrow~~\left\{\! \begin{array}{l} a=17\\b=-56\\c=39 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=56^2-4\cdot 17\cdot 39\\\\ \Delta=3\,136-2\,652\\\\ \Delta=484\\\\ \Delta=22^2$


$y=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\\\\\ y=\dfrac{-(-56)\pm \sqrt{22^2}}{2\cdot 17}\\\\\\ y=\dfrac{56\pm 22}{2\cdot 17}\\\\\\ y=\dfrac{2\cdot 28\pm 2\cdot 11}{2\cdot 17}\\\\\\ y=\dfrac{\diagup\!\!\!\! 2\cdot (28\pm 11)}{\diagup\!\!\!\! 2\cdot 17}$

$y=\dfrac{28\pm 11}{17}\\\\\\ \begin{array}{rcl} y=\dfrac{28-11}{17}&~\text{ ou }~&y=\dfrac{28+11}{17}\\\\ y=\dfrac{17}{17}&~\text{ ou }~&y=\dfrac{39}{17}\\\\ y=1&~\text{ ou }~&y=\dfrac{39}{17} \end{array}$


$\bullet\;\;$ Para $y=1,$

$x=7-4\cdot 1\\\\ x=7-4\\\\ x=3$


$\bullet\;\;$ Para $y=\dfrac{39}{17}\,,$

$x=7-4\cdot \dfrac{39}{17}\\\\\\ x=\dfrac{7\cdot 17}{17}-\dfrac{4\cdot 39}{17}\\\\\\ x=\dfrac{119}{17}-\dfrac{156}{17}\\\\\\ x=\dfrac{119-156}{17}\\\\\\ x=-\,\dfrac{37}{17}$

______________

Então, temos dois pares que satisfazem o sistema:

$\bullet\;\;(x,\,y)=(3,\,1)~~~~\text{ ou }\\\\\\ \bullet\;\;(x,\,y)=\left(-\,\dfrac{37}{17}\,,\;\dfrac{39}{17} \right )$


Bons estudos! :-)