Question

Решить систему уравнений
4x + 3y = 0
x^2 + y^2=100

Answer 1

$\left \{ {{4x+3y=0} \atop {x^2+y^2=100}} \right. \\4x=-3y \\x=- \frac{3}{4}* y \\(- \frac{3}{4}* y)^2+y^2=100 \\ \frac{9y^2}{16} +y^2=100 \\9y^2+16y^2=1600 \\25y^2=1600 \\y^2=64 \\y_1=8 \\y_2=-8 \\x_1=- \frac{3}{4}*8=-6 \\x_2=- \frac{3}{4}*(-8)=6$
Ответ: (-6;8), (6;-8)