Question

iF z = x+y then the value of z is
2
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Answer 1

Answer:

Originally Answered: X^2=y+z, y^2=z+x, z^2 = x+y then the value of 1/1+x + 1/1+y + 1/1+z =?

11+x+11+y+11+z=xx+x2+yy+y2+zz+z2=x+y+zx+y+z=1  

However, a crucial step in this procedure is to ensure that neither of  x,y,z=0 .

If  x=y=z=0  then your answer is  3 .

Suppose  x=0,y=0  then  z=0 . So suppose  x=0,y≠0,z≠0 . Then,  y+z=0,y2=z,z2=y . Subtracting the last to equations,  (y−z)(y+z)=−(y−z)⟹y−z=0⟹y=z . But as  y+z=0  we have  y=z=0 .

So your answer is either  1  or  3 .

Answer 2

Answer:

Originally Answered: X^2=y+z, y^2=z+x, z^2 = x+y then the value of 1/1+x + 1/1+y + 1/1+z =?

11+x+11+y+11+z=xx+x2+yy+y2+zz+z2=x+y+zx+y+z=1

However, a crucial step in this procedure is to ensure that neither of x,y,z=0 .

If x=y=z=0 then your answer is 3 .

Suppose x=0,y=0 then z=0 . So suppose x=0,y≠0,z≠0 . Then, y+z=0,y2=z,z2=y . Subtracting the last to equations, (y−z)(y+z)=−(y−z)⟹y−z=0⟹y=z . But as y+z=0 we have y=z=0 .

So your answer is either 1 or 3 .

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