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

Ответ:

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

Объяснение:

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

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