Question

Solve for $x$ and $y:x$ & $y:x+y+\frac{x}{y}=\frac{1}{2}$ & $(x+y)\frac{x}{y}=\frac{1}{2},x,y$ are real.

Answer 1

The given equations are,

$x+y+\frac{x}{y}=\frac{1}{2}$

$yx+y^2+x=\frac{1}{2}x$

$y^2+xy+x-\frac{1}{2}x=0$

$y^2+xy+\frac{1}{2}x=0$ (1)

And,

$\left(x+y\right)\frac{x}{y}=\frac{1}{2}$

$\frac{x^2}{y}+x=\frac{1}{2}$

$x^2+xy-\frac{1}{2}y=0$ (2)

Subtract equation (2) from equation (1),

$y^2-x^2+\frac{1}{2}\left(x+y\right)=0$

$\left(y+x\right)\left(y-x\right)+\frac{1}{2}\left(x+y\right)=0$

$\left(y+x\right)\left(y-x+\frac{1}{2}\right)=0$

$x+y=0$

$x=-y$

And,

$y-x+\frac{1}{2}=0$

$y+y=-\frac{1}{2}$

$2y=-\frac{1}{2}$

$y=-\frac{1}{4}$

$x=\frac{1}{4}$

Therefore, the values are $x=\frac{1}{4}$ and $y=-\frac{1}{4}$.