Question

развежите систему уравнений:
{х+4у=6
{х^2-3ху-у^2=-3

Answer 1

$\left \{ {{x + 4y = 6} \atop {x^2 - 3xy - y^2 = -3}} \right.$  ⇔ $\left \{ {{x = 6 - 4y} \atop {(x - y)(x + y) - 3xy = -3}} \right.$ ⇔ $\left \{ {{x = 6 - 4y} \atop {(6 - 4y - y)(6 - 4y + y) - 3(6 - 4y)y = -3}} \right.$

⇔$\left \{ {{x = 6 - 4y} \atop {(6 - 5y)(6 - 3y) - 18y + 12y^2 = -3}} \right.$ ⇔ $\left \{ {{x = 6 - 4y} \atop {6(6 - 5y) - 3y(6 - 5y) - 18y + 12y^2 = -3}} \right.$

⇔$\left \{ {{x = 6 - 4y} \atop {36 - 30y - 18y + 15y^2 - 18y + 12y^2 = -3}} \right.$ ⇔ $\left \{ {{x = 6 - 4y} \atop {27y^2 - 66y + 39 = 0}} \right.$ ⇔ $\left \{ {{x = 6 - 4y} \atop {9y^2 - 22y + 13 = 0}} \right.$

⇔$\left \{ {{x = 6 - 4y} \atop {y = 1}} \right.$⇔$\left \{ {{x = 2} \atop {y = 1}} \right.$

   $\left \{ {{x = 6 - 4y} \atop {y = \frac{13}{9}}} \right.$    $\left \{ {{x = \frac{2}{9}} \atop {y = \frac{13}{9}}} \right.$