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Bunuel
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Another way solve it is put
X/(b-c) =Y/(c-a) =z/(a-b) =k
And put the values of x, y, z from above equation in ax+by+cz
It will become 0.
Hence A

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Bunuel
If a, b, and c are three different numbers and \(\frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}\), then what is the value of \(ax + by + cz\) ?

(A) 0

(B) 1

(C) 2

(D) 3

(E) 4

let a=1,b=2 ,c = 3
and x=-1,y=2,z=-1
so \(ax + by + cz\)
IMO A ; 0
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aankit

Hello!

Why are we sure that we can substitute the values with any number?

Kind regards!
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aankit

Hello!

Why are we sure that we can substitute the values with any number?

Kind regards!

as no. of equations is less than no. of variables and if the equation is correct with one or more solution then we can take any value of a,b,c provided they are not equal, these values have to satisfy the equation.
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I have a different solution.

Let x/(b-c)=y/(c-a)=z/(a-b)=1
Hence,
x=b-c
y=c-a
z=a-b

Now lets find out what they want. i.e ax+by+cz
Substituting the values in the above equations we get,
ax=ab-ac
by-bc-ab
cz=ac-bc

Hence ax+by+cz= (ab-ac) + (bc-ab) + (ac-bc) = (ab-ab)+(bc-bc)+(ac-ac)= 0
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