tex]...

July 2022 1 4 Report

решите системы уравнений
[tex]a) \left \{ {{ y^{2}-xy=12 } \atop {3y-x=10}} \right.

b) \left \{ {{3 y^{2}-2xy=28 } \atop {x+3y=-2}} \right. [/tex]

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A)
$\left \{ {{y^2-xy=12} \atop {3y-x=10}} \right.$
$\left \{ {{x=3y-10} \atop {y^2-y(3y-10)=12}} \right.$
$\left \{ {{x=3y-10} \atop {y^2-3y^2+10y-12=0}} \right.$
$\left \{ {{x=3y-10} \atop {-2y^2+10y-12=0}} \right.$
$\left \{ {{x=3y-10} \atop {y^2-5y+6=0}} \right.$
${y^2-5y+6=0$
$D=(-5)^2-4*1*6=1$
$y_1= \frac{5+1}{2} =3,$    $x_1=3*3-10=-1$
$y_2= \frac{5-1}{2} =2,$ $x_2=3*2-10=-4$

Ответ: (-1;3); (-4;2)

b)
$\left \{ {{3y^2-2xy=28} \atop {x+3y=-2}} \right.$
$\left \{ {{x=-2-3y} \atop {3y^2-2y(-2-3y)=28}} \right.$
$\left \{ {{x=-2-3y} \atop {3y^2+4y+6y^2-28=0}} \right.$
$\left \{ {{x=-2-3y} \atop {9y^2+4y-28=0}} \right.$
$9y^2+4y-28=0$
$D=4^2-4*9*(-28)=1024$
$y_1= \frac{-4+32}{18} = \frac{14}{9}= 1 \frac{5}{9},$    $x_1=-2-3* \frac{14}{9} =-6 \frac{2}{3}$
$y_2= \frac{-4-32}{18} =-2,$     $x_2=-2-3*(-2)=4$

Ответ: $(-4 \frac{2}{3};1 \frac{5}{9} )$ ; $(4;-2)$

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