Question

If $y\ne 0$, then the number of values of the pair $(x,y)$ such that $x+y+\frac{x}{y}=\frac{1}{2}$ and $(x+y)\frac{x}{y}=-\frac{1}{2},$ is

• A. $1$
• B. $2$
• C. $0$
• D. none of these

Answer 1

The correct option is B $2$

By using quadratic $x+y$ as one root and $x/y$ as another root we can get the roots as $1,-1/2$
If $x/y=1$ and $x+y=-1/2$ then possible values are $x=y=-1/4$
If $x/y=-1/2;x+y=1$ then $x=-1,y=2$
So total no of pairs are $2$