if x + y + z =3, x +2y +3z =4 ,x + 4y + 9z =6, then (y,z)=
x + y + z =3 ........(let this be equation 1)
x +2y +3z =4........(let this be equation 2)
x + 4y + 9z =6........(let this be equation 3)
(y,z) = (1,0)
From equation 1 (we get
x = 3-y-z
Substitute value of x equation 2 & 3 we get:
y + 2z = 1.........(let this be equation 4)
3y =3
respectively.
Hence
y=3/3
substituting value of y in equation 4 we get z =0
HENCE
y=1
z=0
(y,z) = (1,0)