Решите систему уравнений
(x+2)^2+(y-1)^2 = x^2+y^2+5
3*y+x = 28
(x+2)^2+(y-1)^2=x^2+y^2+5
3y+x=28
3y+x=28x=-3y+28 (x+2)^2+(y-1)^2=x^2+y^2+5
x^2+4x+4+y^2-2y+1-x^2-y^2-5=0
4x+5+y^2-2y-y^2-5=0
4x-2y=04*(-3y+28)-2y=0 -14y+112=0
y=112/14
y=83y+x=283*8+x=28
x=4
(x+2)^2+(y-1)^2=x^2+y^2+5, 3*y+x=28
x = 4, y = 8;
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Answers & Comments
(x+2)^2+(y-1)^2=x^2+y^2+5
3y+x=28
3y+x=28
x=-3y+28
(x+2)^2+(y-1)^2=x^2+y^2+5
x^2+4x+4+y^2-2y+1-x^2-y^2-5=0
4x+5+y^2-2y-y^2-5=0
4x-2y=0
4*(-3y+28)-2y=0
-14y+112=0
y=112/14
y=8
3y+x=28
3*8+x=28
x=4
(x+2)^2+(y-1)^2=x^2+y^2+5, 3*y+x=28
x = 4, y = 8;