Question

pe=y
and
x+y=1
then
(x, y) =
​

Answer 1

Answer:

Given that xyz=1

⟹y−1=xz and ⟹y=(xz)−1 .

The key idea is to write the y in terms of x,z .

Now,

1(1+x+y−1)+1(1+y+z−1)+1(1+z+x−1)

=1(1+x+xz)+1(1+(xz)−1+z−1)+1(1+z+x−1)

=1(1+x+xz)+xz(xz+1+x)+x(x+xz+1)

=1(1+x+xz)+xz(1+x+xz)+x(1+x+xz)

=1+x+xz(1+x+xz)

=1

∴1(1+x+y−1)+1(1+y+z−1)+1(1+z+x−1)=1

Note : You can select any one variable among x,y,z and write interms of two others

Answer 2

Answer:

$\huge\underline\bold\blue{AnsWeR}$

$xyz = 1 xy = 1/z, yz = 1/x , xz = 1/y$