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TheUltimateWinner
If y ≠ z, is \(\frac{(x-y)+(z-y)}{z-y}=2\)?


(1) x - y = 2
(2) x = z


The first step in such questions is to simplify the equation.

\(\frac{(x-y)+(z-y)}{z-y}=2\)
\((x-y)+(z-y)=2(z-y)........x+z-2y=2z-2y.......x=z\)

So the question finally becomes -- Is x=z?

(1) x - y = 2
Nothing about z

(2) x = z
exact information that we are looking for.

B
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Hello!
In statment 1
2+Z-Y/(Z-Y) = 2
2+Z-Y = 2Z-2Y
2+Y = Z
z-y=2, SO ((x-y)+(z-y))/(z-y) = (2+2)/2 = 2 Suf! No?
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Hello!
From statment 1:"x-y=2"
(2+(z-y))/(z-y)
2+z-y=2z-2y
"z-y=2".... SO (2+2)/2=2.... SUF! No?
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