Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 1: $0$
Option 2: $1$
Option 3: $2$
Option 4: $–1$
Correct Answer: $2$
Solution :
Given:
$\frac{(x+y)}{z}=2$
⇒ $y=2z-x$
⇒ $y-z=z-x$
Now, $[\frac{y}{(y-z)}]+[\frac{x}{(x-z)}]$
= $[\frac{y}{(z-x)}]+[\frac{x}{(x-z)}]$
= $[-\frac{y}{(x-z)}]+[\frac{x}{(x-z)}]$
= $[\frac{-y+x}{(x-z)}]$
= $[\frac{-(2z-x)+x}{(x-z)}]$ [After putting value of $y$]
= $[\frac{2(x-z)}{(x-z)}]$
= $2$
Hence, the correct answer is $2$.
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