1. Решить систему уравнений методом подстановки:а) x+y=3, y ² - xy = -1 ;б) 1/х + 1/у...

1.
а)
$\left \{ {{x+y=3} \atop {y^2 - xy = -1}} \right.\\ \left \{ {{x=3-y} \atop {y^2 - (3-y)y = -1}} \right.\\ y^2 - 3y+y^2+1=0\\ 2y^2-3y+1=0\\ D=9-8=1\\ y_1=\frac{3+1}{4}=1\\ y_2=\frac{3-1}{4}=\frac{1}{2}\\ x_1=3-y_1=3-1=2\\ x_2=3-y_2=3-\frac{1}{2}=2\frac{1}{2}\\$
Ответ: $(2;1), (2\frac{1}{2};\frac{1}{2})$

б)
$\left \{ {{1/x + 1/y =3/4} \atop {x-y=2}} \right. \\ x=2+y\\ \frac{1}{2+y}+\frac{1}{y}=\frac{3}{4}\\ \frac{y+(2+y)}{y(2+y)}=\frac{3}{4}\\ \frac{2y+2}{2y+y^2}=\frac{3}{4}\\ 4(2y+2)=3(2y+y^2)\\ 8y+8=6y+3y^2\\ 3y^2-2y-8=0\\ D=4+96=100\\ y_1=\frac{2+10}{6}=2\\ y_2=\frac{2-10}{6}=\frac{4}{3}=1\frac{1}{3}\\ x_1=y_1+2=2+2=4\\ x_2=y_2+2=\frac{4}{3}+2=3\frac{1}{3}$
Ответ: $(4; 2), (3\frac{1}{3}; 1\frac{1}{3})$

2.
$\left \{ {{2x^2 + 3y^2=14} \atop {-x^2 +2y^2 = 7}} \right. \\ \left \{ {{2x^2 + 3y^2=14} \atop {-2x^2 +4y^2 = 14}} \right. \\ 7y^2 = 28\\ y^2 = 4\\ y_1 = 2\\ y_2=-2\\ \\ \\ y=2\\2x^2 + 3*2^2=14\\ 2x^2 + 12=14\\ 2x^2 =2\\ x^2=1\\ x_1=1\\ x_2=-1\\ \\ \\ y=-2\\2x^2 + 3*(-2)^2=14\\ 2x^2 + 12=14\\ 2x^2 =2\\ x^2=1\\ x_1=1\\ x_2=-1\\$
Ответ: $(1; 2), (-1; 2), (1; -2), (-1; -2)$