Question

Resolva os sistemas de equações:
COM EXPLICAÇÃO

Answer 1

    $\left \{ {{x=2y} \atop {x+y^2=35}} \right. \Leftrightarrow\left \{ {{x=2y} \atop {2y+y^2=35}} \right. \Leftrightarrow\left \{ {{x=2y} \atop {y^2+2y-35=0}} \right.\Leftrightarrow$

$\Leftrightarrow\left \{ {{x=2y} \atop {y=-7}} \right.\;\vee\;\left \{ {{x=2y} \atop {y=5}} \right.\Leftrightarrow\left \{ {{x=2\times(-7)} \atop {y=-7}} \right.\;\vee\;\left \{ {{x=2\times5} \atop {y=5}} \right.\Leftrightarrow\left \{ {{x=-14} \atop {y=-7}} \right.\;\vee\;\left \{ {{x=10} \atop {y=5}} \right.$

Resposta:  $(x\;;\;y)=(-14\;;\;-7)\;\vee\;(10\;;\;5)$

Cálculos Auxiliares

    $y=\frac{-2\pm\sqrt{2^2-4\times1\times(-35)}}{2\times1}\Leftrightarrow$

$\Leftrightarrow y=\frac{-2\pm\sqrt{4-4\times(-35)}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-2\pm\sqrt{4+140}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-2\pm\sqrt{144}}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-2\pm12}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-2-12}{2}\;\vee\;y=\frac{-2+12}{2}\Leftrightarrow$

$\Leftrightarrow y=\frac{-14}{2}\;\vee\;y=\frac{10}{2}\Leftrightarrow$

$\Leftrightarrow y=-7\;\vee\;y=5$