Maths about ratio

[fork]

1)If (x+y-z)/3
=(y+z-x)/4
=(z+x-y)/5,
then x:y:z=
A.3:4:5
B.8:7:9
C.9:16:25
D.15:12:20

The answer for the above question is option B. But I don't know how to solve this kind of questions. Can anyone tell me what should I do in the first step? Thanks for answering my question.

[AntonVrba]

step 1: 3*4*5=60
step 2: 20x+20y-20z=15y+...

[fork]

Step2:20x+20y-20z=15y+15z-15x=12z+12x-12z
Step3:35x+32y=27y+35z=32z+27x

x=4(z-y)
y=7(z-x)
z=9(x-y)

That's what I solve from the above equation. After that, what should I do? Thanks.

[HallsofIvy]

Did you see AntonVrba's response? Since you asked to choose among different possible answers, the simplest thing to do is try each.
x= 3, y= 4, z= 5 ARE in ratio 3:4:5. Do those numbers satisfy the original equations?

If not try x= 8, y=7, z= 9, etc.

[AntonVrba]

no need for trial and error of trying given posibilities, you can calculate
from the three relationships x,y,z that you already have solved by substituting first y and then for second z in the first of your equations.

x = 4z - 4y
= 4z - 28z + 28x
or
24z = 27x
8z = 9 x
8/9 = x/z or 8:9 = x:z

and

x = 4z - 4y
x = 36x - 36y - 4y
........

[HallsofIvy]

Yeah, but it works and thought it would be simpler for fork.