Question

Решите способом сложения систему уравнений x+3y=11 и 2x+y^2=14​

Answer 1

$\left \{ {{x+3y=11} \atop {2x+y^2=14}} \right.$

$\left \{ {{-2x-6y=-22} \atop {2x+y^2=14}} \right.$

$y^2-6y=-8$

$y^2-6y+8=0$

D $= b^2-4ac = 36-32=4$

$y_{1} =\frac{6+2}{2}=4$

$y_{2} =\frac{6-2}{2} =2$

$\left \{ {{x+3y=11} \atop {y_{1}=4}} \right.$

$\left \{ {{12+x=11} \atop {y_{1}=4}} \right.$

$\left \{ {{y_{1}=4 } \atop {x_{1}=-1}} \right.$

$\left \{ {{x+3y=11} \atop {y_{2}=2}} \right.$

$\left \{ {{x+6=11} \atop {y_{2}=2}} \right.$

$\left \{ {{y_{2}=2} } \atop {x_{2}=5}} \right.$

Ответ: $(-1;4); (5;2 )$