Is it too what?

Algebra Level pending

{ ( 1 + x ) ( 1 + y ) = 4 y ( 1 + y ) ( 1 + z ) = 4 z ( 1 + z ) ( 1 + x ) = 4 x \large \begin{cases} (1+x)(1+y)=4y\\ (1+y)(1+z)=4z \\ (1+z)(1+x)=4x \end{cases}

Let x , y x,y and z z be numbers satisfying the equation above, find x + y + z x+y+z .


The answer is 3.

Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Ayush G Rai
May 12, 2016

Let x = u + 1 , y = v + 1 x=u+1,y=v+1 and z = w + 1 z=w+1 .From he first equation,we get
( u + 2 ) ( v + 2 ) = 4 ( v + 1 ) (u+2)(v+2)=4(v+1) ;when expand and simplify you get u v = 2 v − 2 u . − − − 1 uv=2v-2u.- - - 1
Similarly, v w = 2 w − 2 v − − − 2 vw=2w-2v - - - 2
and w u = 2 u − 2 w . − − − 3 wu=2u-2w. - - - 3


Hence,
u v w = 2 v w − 2 w u uvw=2vw-2wu , [when we multiply the 1st equation by w]
u v w = 2 w u − 2 u v uvw=2wu-2uv , [when we multiply the 2nd equation by u]
u v w = 2 u v − 2 v w . uvw=2uv-2vw. [when we multiply the 3rd equation by v]

Adding these three relations we get 3 u v w = 0 3uvw=0 .
Suppose u = 0 ; x = u + 1 ⇒ x = 0 + 1 ⇒ x = 1 u=0;x=u+1\Rightarrow x=0+1\Rightarrow x=1 .
In the first equation of our problem ( 1 + x ) ( 1 + y ) = 4 y (1+x)(1+y)=4y .
Substitute x = 1 x=1 in the first equation.
We get, ( 1 + 1 ) ( 1 + y ) = 4 y (1+1)(1+y)=4y
( 2 ) ( 1 + y ) = 4 y (2)(1+y)=4y
( 1 + y ) = 2 y (1+y)=2y
y = 1 y=1 .
Again if we substitute y = 1 y=1 in the second equation we get z = 1 z=1 .
So, x + y + z = 1 + 1 + 1 = 3 x+y+z=1+1+1=\boxed 3 .

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