Решите систему уравнений 1)x+y=3 xy=-10 2) x2-y2=5 2x+y=4

[latex]x+y=3, \ xy=-10\\ x=3-y, \\(3-y)y=-10\\ 3y-y^2+10=0\\ y^2-3y-10=0\\ D=3^2-4*(-10)=9+40=49\\ y_1= \frac{3+7}{2} =5\\ y_2= \frac{3-7}{2} =-2\\ x_1=3-5=-2\\ x_2=3-(-2)=5\\ Answer: \ (-2;5), \ (5;-2)[/latex] [latex]x^2-y^2=5, \ 2x+y=4\\ y=4-2x\\ x^2-(4-2x)^2=5\\ x^2-16+16x-4x^2-5=0\\ -3x^2+16x-21=0\\ 3x^2-16x+21=0\\ D=16^2-4*3*21=256-252=4\\ x_1= \frac{16+2}{6} =3\\ x_2= \frac{16-2}{6} =2 \frac{1}{3}\\ y_1=4-2*3=-2\\ y_2=4-2*2 \frac{1}{3} =4-4 \frac{2}{3} =- \frac{2}{3}\\ Answer: \ (3;-2), \ (2 \frac{1}{3};- \frac{2}{3})[/latex]