x = 2y + 1
y(2y+ 1) + y = 12
x = 2y + 1
2y^2 + y + y = 12
x = 2y + 1
2y^2 + 2y - 12 = 0 *
*
2y^2 + 2y - 12 = 0 // : 2
y^2 + y - 6 = 0
D = 1 + 4*6 = 25 = 5^2
y1 = ( - 1 + 5)/2 = 4/2 = 2;
y2 = ( - 1 - 5)/2 = - 6/2 = - 3;
x1 = 2y + 1 = 2*2 + 1 = 5
y1 = 2
x2 = 2*(-3) + 1 = - 5
y2 = - 3
Ответ:
( - 5; 3); ( 5; 2)