Question

10. Resolver o sistema de equações x + 2y = 7 e y2 - x2 = 8.

Answer 1

$\boxed{\begin{array}{l}\begin{cases}\sf x+2y=7\\\sf y^2-x^2=8\end{cases}\\\begin{cases}\sf x=7-2y\\\sf y^2-x^2=8\end{cases}\\\sf y^2-(7-2y)^2=8\\\sf y^2-(49-28y+4y^2)-8=0\\\sf y^2-49+28y-4y^2-8=0\\\sf -3y^2+28y-57=0\cdot(-1)\\\sf 3y^2-28y+57=0\end{array}}$

$\boxed{\begin{array}{l}\sf\Delta=b^2-4ac\\\sf\Delta=(-28)^2-4\cdot3\cdot57\\\sf\Delta=784-684\\\sf\Delta=100\\\sf y=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf y=\dfrac{-(-28)\pm\sqrt{100}}{2\cdot3}\\\sf y=\dfrac{28\pm10}{6}\begin{cases}\sf y_1=\dfrac{28+10}{6}=\dfrac{38\div2}{6\div2}=\dfrac{19}{3}\\\sf y_2=\dfrac{28-10}{6}=\dfrac{18}{6}=3\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\sf quando~y=\dfrac{19}{3}:\\\sf x=7-2y\\\sf x=7-2\cdot\dfrac{19}{3}\\\sf x=\dfrac{21-38}{3}=-\dfrac{17}{3}\\\sf quando~y=3\\\sf x=7-2y\\\sf x=7-2\cdot3\\\sf x=7-6\\\sf x=1\\\sf S=\bigg\{\bigg(-\dfrac{17}{3},\dfrac{19}{3}\bigg),(1,3)\bigg\}\end{array}}$