Question

Sa se rezolve sistemele de ecuatiii
a ) x+y=3 si xy= -10 b ) x+y= -3 si xy=5supra4 c) x+y=0 xy= - 2
Va roggg ajutati-ma

Answer 1

$\left \{ {{x+y=3} \atop {xy=-10}} \right. =\ \textgreater \ \left \{ {{x=3-y} \atop {(3-y)*y=-10}} \right. =\ \textgreater \ \left \{ {{x=3-y} \atop {3y-y^{2}=-10}} \right. =\ \textgreater \ \left \{ {{x=3-y} \atop {-y^{2}+3y+10=0}} \right. \\ \\ -y^{2}+3y+10 = 0 \\ delta=(+3)^{2}-4*(-1)*(+10)= 9+40 = 49 \ \textgreater \ 0\\ y_1= \frac{-3-\sqrt{delta}}{2*(-1)}=\frac{-3-7}{-2}=5\\ x_1=3-y_1 = 3-5 = -2\\ \\ y_2=\frac{-3+ \sqrt{delta}}{2*(-1)} =\frac{7-3}{-2}=-2\\ x_2=3-y_2=3-(-2)=5 \\ \\ S=(-2; 5) ; (5; -2) \\ \\ \\ \left \{ {{x+y=-3} \atop {xy=\frac{5}{4}}} \right. =\ \textgreater \ \left \{ {{x=-(3+y)} \atop {-4[y*(3+y)]=5}} \right. =\ \textgreater \ \left \{ {{x=-(3+y)} \atop {-4(y^{2}+3y)=5}} \right. =\ \textgreater \ \left \{ {{x=-(3+y)} \atop {-4y^{2}-12y-5=0} \right. \\ \\ -(4y^{2}+12y+5)= 0\\ delta=12^{2}-4*4*5= 144-80=64\ \textgreater \ 0\\ y_1=\frac{-12- \sqrt{delta}}{2*4}=\frac{-12-8}{8}=\frac{-5}{2}\\ x_1= -3+\frac{5}{2} = \frac{-1}{2}\\ \\ y_2=\frac{-12+ \sqrt{delta} }{2*4}=\frac{8-12}{8}=\frac{-1}{2} =\ \textgreater \ x_2 = \frac{-5}{2} \\ \\ \\ \left \{ {{x+y=0} \atop {xy=-2}} \right. =\ \textgreater \ \left \{ {{x=-y} \atop {-y^{2}=-2}} \right. =\ \textgreater \ \left \{ {{x=-y} \atop {y^{2}=2}} \right. \\ \\ y_1=-\sqrt{2} =\ \textgreater \ x_1= \sqrt{2}\\ y_2=+\sqrt{2} =\ \textgreater \ x_2=-\sqrt{2}$

Answer 2

a)

[tex]\\x=3-y\\ (3-y)y=-10\\ 3y-y^2=-10\\ y^2-3y-10=0\\ y^2-5y+2y-10=0\\ y(y-5)+2(y-5)=0\\ (y-5)(y+2)=0\\ y-5=0, y=5, x=-2\\ y+2=0, y=-2, x=5[/tex]
b)

[tex]\\x=-y-3\\ (-y-3)y=\frac{5}{4}\\ -y^2-3y=\frac{5}{4}\\ 4y^2+12y+5=0\\ 4y^2+2y+10y+5=0\\ 2y(2y+1)+5(2y+1)=0\\ (2y+1)(2y+5)=0\\ y=-\frac{1}{2}, x=-\dfrac{5}{2}\\ y=-\frac{5}{2}, x=-\frac{1}{2}[/tex]
c)

[tex]\\x=-y\\ -y^2=-2\\ y^2=2\\ y=\sqrt{2}, x=-\sqrt{2}\\ y=-\sqrt{2}, x=\sqrt{2}[/tex]