A glance could give the solution
If a x = b y = c z for distinct values of a , b , c with a b c = 1 , then find the value of x y + y z + x z .
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
It must be mentioned that x , y , z = 0 .
Log in to reply
Ya also that a,b,c are distinct.
sorry for the inconvenience , @Nihar Mahajan i just forgot to mention that because i was busy in some work so i just put up the question quickly ...
I'm sorry, but if a = b = c = 1 and also x = y = z = 1 , I dont see how 1 + 1 + 1 = 0
Or generally if a = b = c = 1 then any value of x,y, and z satisfies this equation.
Log in to reply
So I think that the problem poser must mention that a , b , c are distinct .
I also made the same mistake and this question is totally flawed.
Nice solution!
Good solution !!
You assumed a x + b y + c z = k and then implied that a = k x 1 , b = k y 1 , c = k z 1 . But if yo substitute the later into the former you get 3 k = k .
can we put K=1 if yes than (xy + yz+xz)/xyz= 1
I just said: imagine if a is 1, and b and c are inverses of each other, so the product of all 3 give 1. Then if x y and z is 0. Well they all give 1. so xy+yz+xz is 0
Let a x = b y = c z = k
⇒ a = k x 1 , b = k y 1 , c = k z 1
⇒ a b c = k x 1 + y 1 + z 1 = k x y z x z + x y + z y = 1
⇒ x y z x z + x y + z y = 0
∴ x z + x y + z y = 0