If x = frac{a}{a-1} and y = frac{1}{a-1},then | vedclass

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(D) Given $x = \frac{a}{a-1}$ and $y = \frac{1}{a-1}$.We can rewrite $x$ as:$x = \frac{a-1+1}{a-1} = \frac{a-1}{a-1} + \frac{1}{a-1} = 1 + \frac{1}{a-

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$x > y$

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(D) Given $x = \frac{a}{a-1}$ and $y = \frac{1}{a-1}$.We can rewrite $x$ as:$x = \frac{a-1+1}{a-1} = \frac{a-1}{a-1} + \frac{1}{a-1} = 1 + \frac{1}{a-1}$.Since $y = \frac{1}{a-1}$,we substitute $y$ into the equation for $x$:$x = 1 + y$.Since $x$ is $1$ greater than $y$,it follows that $x > y$ for all values of $a$ where the expressions are defined $(a 
eq 1)$.

If $x = \frac{a}{a-1}$ and $y = \frac{1}{a-1}$,then

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