If a/x=b/y=c/z and abc=1,show x+y+z=0

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The problem provides the information that a`/x=b/y=c/z` and `abc=1` , hence, using properties of ratios yields:

`a/x=b/y=c/z =>(abc)/(xbc)=(abc)/(yac)=(abc)/(zab) => 1/(xbc)= 1/(yac) = 1/(zab) => xbc = yac = zab => {(x = y*(a/b)),(z = y*(c/b)):}`

You need to test if `x + y + z = 0` , hence, replacing `y*(a/b)` for x and `y*(c/b)` for z yields:

`y*(a/b) + y + y*(c/b) = 0`

Factoring out y yields:

`y(a/b + 1 + c/b) = 0 => y((a + b + c)/b) = 0`

` abc = 1, => (a + b + c)/b != 0 => y = 0`

Replacing 0 for y yields:

`x = y*(a/b) => x = 0*(a/b) => x = 0`

`z = y*(c/b) => z = 0*(c/b) => z = 0`

Since the given ratios `a/x=b/y=c/z` are valid for `x,y,z !=0` , hence, testing if `x + y + z = 0` , under the given conditions, yields `x + y + z != 0.`

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Let

`a/x=b/y=c/z=r`

also `r>0 .`

`` So we have

`a=rx`

`b=ry`

`c=rz`

Given     `abc=1`  ,this means

`rx.ry.rz=1`

`r^3(xyz)=1`

`r^3!=0 ,and xyz!=0`

`` This implies nether x nor y nor z is zero.

aslo `x ,y,z >0`

`x+y+z!=0`

so x+y+z=0 not possible.

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