Question

Решите систему уравнений x+y=3 x^2+y^2=29

Answer 1

{x+y=3

{x^2+y^2=29

x=3-y

(3-y)^2+y^2=29

9-6y+y^2+y^2=29

2y^2-6y-20=0  | 2

y^2-3y-10=0

y₁+y₂=3

y₁*y₂=-10

y₁=5

y₂=-2

x₁=3-5=-2

x₂=3-(-2)=5

Answer 2

(5;-2),(-2;5).

Объяснение:

$\left \{ \begin{array}{lcl} {{x+y=3,} \\ {x^{2}+y^{2} =29;}} \end{array} \right\Leftrightarrow\left \{ \begin{array}{lcl} {{y=3-x,} \\ {x^{2}+(3-x)^{2} =29;}} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=3-x,} \\ {x^{2}+9-6x+x^{2} -29=0; }} \end{array} \right.\Leftrightarrow$

$\left \{ \begin{array}{lcl} {{y=3-x,} \\ {x^{2}-3x-10=0; }} \end{array} \right.\Leftrightarrow\left \{ \begin{array}{lcl} {{y=3-x,} \\\left [ \begin{array}{lcl} {{x=-2} \\ {x=5;}} \end{array} \right {}} \end{array} \right\Leftrightarrow\left [\begin{array}{lcl} {\left \{ \begin{array}{lcl} {{x=5} \\ {y=-2}} \end{array} \right.{} \\ {\left \{ \begin{array}{lcl} {{x=-2} \\ {y=5}} \end{array} \right.}} \end{array} \right.$