Question

Решить систему уравнений: 1) 4x^2-y=2 3x-2y=-1 2) x/y=1/3 x^2+y^2=50

Answer 1

$\left \{ {{4 x^{2} -y=2} \\ \atop {3x-2y=-1}} \right. \\ \\ \left \{ {{y=4 x^{2} -2} \\ \atop {3x-2(4 x^{2} -2)=-1}} \right. \\ \\ \left \{ {{y=4 x^{2} -2} \\ \atop {3x-8 x^{2} +4+1=0}} \right. \\ \\ \left \{ {{y=4 x^{2} -2} \\ \atop {8 x^{2} -3x-5=0}} \right. \\ \\ \\ D=3^2+4*8*5=169=13^2 \\ \\ x_1= \frac{3+13}{2*8}=1 \\ \\ x_2= \frac{3-13}{2*8} =-\frac{5}{8} =-0.625 \\ \\ y_1=4*1-2=2 \\ \\ y_2= 4*(- \frac{5}{8} )^2-2=- \frac{7}{16} =-0,4375$

ответ (1;  2); (-0,625; -0,4375)

$\left \{ {{ \frac{x}{y} = \frac{1}{3} } \atop { x^{2} +y^2=50}} \right. \\ \\ \left \{ {{y=3x} \atop {x^2+y^2=50}} \right. \\ \\ \left \{ {{y=3x} \atop {x^2+(3x)^2=50}} \right. \\ \\ \left \{ {{y=3x} \atop {x^2+9x^2=50}} \right. \\ \\ \left \{ {{y=3x} \atop {10x^2=50}} \right. \\ \\ 10 x^{2} =50 \\ \\ x^{2} =5 \\ \\ x_1= \sqrt{5} \\ \\ x_2=- \sqrt{5} \\ \\ y_1= 3 \sqrt{5} \\ \\ y_2=-3 \sqrt{5}$

ответ (√5; 3√5), (-√5; -3√5)