Question

Resolva:

$a)\left \{ {{2x-y^{2} =1} \atop {3x+y=4}} \right. \\\\b) \left \{ {{x+6y=3} \atop {x.y=-9}} \right. \\\\c)\left \{ {{x-y=8} \atop {x+y^{2} =14}} \right. \\\\d) \left \{ {{x+y=\frac{5}{6}} \atop {x^{2}+ 3y=2}} \right.$

Me ajudem, por favor.

Answer 1

Resposta:

$a)\\ \\ 2x-y^{2} =1~~equacao~1\\ \\ 3x+y=4~~equacao~2\\ \\ eq~1:\\ \\ 2x-y^{2}=1\\ \\ 2x=1+y^{2} \\ \\ x=\frac{1+y^{2} }{2} \\ \\ eq~2:\\ \\ 3x+y=4\\ \\ 3x=4-y\\ \\ x=\frac{4-y}{3} \\ \\ x=x~~logo:\\ \\ \frac{1+y^{2} }{2} =\frac{4-y}{3} \\ \\ 3(1+y^{2})=2(4-y)\\ \\ 3y^{2} +3=8-2y\\ \\ 3y^{2} +2y+3-8=0\\ \\ 3y^{2} +2y-5=0\\ \\ delta=b^{2}-4.a.c\\  \\ delta=2^{2} -4.3.(-5)\\ \\ delta=4+60\\ \\ delta=64\\ \\ y=\frac{-b+-\sqrt{delta} }{2.a} \\ \\ y=\frac{-2+-\sqrt{64} }{6}$

$y=\frac{-2+-8}{6} \\ \\ y'=\frac{-2-8}{6} =-\frac{10}{6} \\ \\ y''=\frac{-2+8}{6} \\ \\ y''=\frac{6}{6} \\ \\ y''=1\\ \\ substitua~y=1~em~uma~das~equacoes:\\ \\ eq~1:\\ \\ 2x-y^{2} =1\\ \\ 2x-1^{2} =1\\ \\ 2x-1=1\\ \\ 2x=1+1\\ \\ 2x=2\\ \\ x=\frac{2}{2} \\ \\ x=1\\ \\ S~~(x=1;~y=1;~x=y)$

$b)\\ \\ x+6y=3~~eq~1\\ \\ xy=-9~eq~2\\ \\ eq~1:\\ \\ x+6y=3\\ \\ x=3-6y\\ \\ eq~2\\ \\ x.y=-9\\ \\ x=\frac{-9}{y} \\ \\ x=x~~logo:\\ \\ 3-6y=\frac{-9}{y} \\ \\ y(3-6y)=-9\\ \\ -6y^{2}+3y+9=0\\  \\ delta=b^{2} -4.a.c\\ \\ delta=9-4.(-6).9\\ \\ delta=9+216\\ \\ delta=225\\ \\ y=\frac{-b+-\sqrt{delta} }{2.a} \\ \\ y=\frac{-3+-\sqrt{225} }{-12} \\ \\ y=\frac{-3+-15}{-12} \\ \\ y'=\frac{-3-15}{-12} =\frac{-18}{-12} =\frac{3}{2} \\ \\ y''=\frac{-3+15}{-12} \\ \\ y''=\frac{12}{-12}$

$y''=-1\\ \\ substitua~(y=-1)~em~uma~das~equacoes:\\ \\ eq~1:\\ \\ x+6y=3\\ \\ x+6(-1)=3\\ \\ x-6=3\\ \\ x=3+6\\ \\ x=9\\ \\ S~~(x=9;~~y=-1)$