Resposta:
Se [tex]y=-6[/tex], [tex]x=-2,5[/tex]
Se [tex]y=5[/tex], [tex]x=3[/tex]
Explicação passo-a-passo:
[tex]\left \{ {{2x+y^2=31} \atop {2x-y=1}} \right. \\\left \{ {{2x+y^2=31} \atop {-2x+y=-1}} \right. \\\\y^2+y=30\\y^2+y-30=0[/tex]
Δ [tex]= 1^2-4.(-30)[/tex]
Δ [tex]=1+120[/tex]
Δ [tex]= 121[/tex]
Δ [tex]= 11^2[/tex]
[tex]y_{1}=\frac{-1-11}{2}\\y_1=\frac{-12}{2}\\y_1= -6\\\\y_{2}=\frac{-1+11}{2}\\y_2= \frac{10}{2}\\y_2= 5[/tex]
Para [tex]y=-6[/tex]
[tex]2x+6=1\\2x=1-6\\2x=5\\x=-\frac{5}{2}\\x=-2,5[/tex]
Para [tex]y=5[/tex]
[tex]2x-5=1\\2x=1+5\\2x=6\\x=\frac{6}{2}\\x=3[/tex]
