Question

If $x<0,y<0,x+y+\frac{x}{y}=\frac{1}{2}$ and $(x+y)\frac{x}{y}=-\frac{1}{2}$, then y= ......

Answer 1

Let x < 0 , y < 0

given $x+y+\frac{x}{y}=\frac{1}{2}...eq.1\&(x+y)\frac{x}{y}=\frac{-1}{2}...eq.2$

Let $x+y=a,\frac{x}{y}=b$ put in eq.1 & eq.2

$\Rightarrow a+b=\frac{1}{2}.,..eq.3$

$\Rightarrow ab=\frac{-1}{2}.,..eq.4$

$\Rightarrow b=\frac{-1}{2a}$ Put in eq.3

$\Rightarrow a+\frac{-1}{2a}=\frac{1}{2}$

$\Rightarrow \frac{2a^2-1}{2a}=\frac{1}{2}$

$\Rightarrow 2a^2-1=a\Rightarrow 2a^2-a-1=0\Rightarrow (a-1)(2a+1)=0\Rightarrow a=\frac{-1}{2}ora=-1$

when $a=1\Rightarrow b=\frac{-1}{2}$

$\Rightarrow x+y=1,\frac{x}{y}=b\Rightarrow \frac{x}{y}=\frac{-1}{2}\Rightarrow y=-2x$

$x-2x=1\Rightarrow x=-1,y=-2\Rightarrow$ Not satisfied $x+y=1$

when $a=\frac{-1}{2},b=1$

$x+y=\frac{-1}{2},\frac{x}{y}=1\Rightarrow x=y$

$\Rightarrow 2x=\frac{-1}{2}x=\frac{-1}{4},y=\frac{-1}{4}$

$\therefore$ These is right answer because x < 0 , y < 0