(1) solve:x^2 -xy=6, xy+y^2=4?

Question

(2) If x=cy+bz, y=az+cx, z=bx+ay,where x,y,z are not all zero,prove that a^2+b^2+c^2+2abc=1.Also prove that in this case
x^2/(1-a^2)=y^2/(1-b^2)=z^2/(1-c^2)

please write step by step

Answer 1

add

x^2+y^2 = 10

multiply 1st by 4 and 2nd by 6
4(x^2-xy) = 6(y^2+xy)
or 4x^2-10xy-6y^2 = 0
2x^2-5xy-3y^2 = 0
(x-3y)(2x+y) = 0

x = 3y or y = 2x

x^2+y^2 = 10
x= 3y
=> 10y^2 = 10

y = 1 x = 3 or y = -1 x =-3

x^2+y^2 = 10
y= 2x -> x= sqrt(2) y = 2sqrt(2) or x = -sqrt(2) y = - 2sqrt(2)

Answer 2

1) x = 3
y = 1

Answer 3

sum x^2-xy+xy+y^2=10 then x^2+y^2=10 then this equation of a circle the coordinate center (0 , 0) wirh the radius r=sqrt(10)