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1 solution
we can solve this by cross multiplying so we get x(c+a-b)=z(c+a-b)
since all of them are equal to each other you can cross multiply 2 numerators by the denominator of the third one
ex: a/b=c/d=e/f so af=cf
if we divide both equations by (c+a-b) we get x=z
if we repeat the process with y and z we get y=z ,therefore x=y=z
if the numerators are equal then the denominators must be equal
ex: 5/a =5/b so 5a=5b ,therefore a=b
so b+c-a=c+a-b b+c=c+2a-b b=2a-b 2b=2a a=b we can conclude from that that c=a=b
so b-c(x)+c-a(x)+a-b(x)=c-c(x)+c-c(x)+c-c(x)
=/boxed{0}
sorry for any confusion my solution caused
I'm a bad teacher (: