Plus Minus Times Divide

From LHS: ${y}$ = $\frac{1}{y}$ $\Rightarrow$ $y^{2}$ = 1 $\Rightarrow$ ${y}$ = $\pm$ 1. Now if ${y}$ = 1, then we have ${x}$ = ${x}$ = ${x}$ +1, where the RHS results in 0=1, so by contradiction ${y}$ = -1, $\Rightarrow$ - ${x}$ = ${x}$ -1 $\Rightarrow$ ${x}$ = $\frac{1}{2}$ .

$\therefore$ ${x}$ - ${y}$ = $\boxed{\frac{3}{2}}$

We've got x,y could not be 0 => xy=x/y <=> y=1/y => y= 1 or -1. If y=1 =>x=x+1 <=> 0=1 so y must be -1 => -x=x-1 <=> x=1/2 => x-y=1/2-(-1)=3/2=1.5

If xy=x/y, then x*y^2=x, therefore y^2=1.

Now, if x/y=x+y, then x=xy+y^2, so x=xy+1, and 1=y+1/x, therefore y=1-1/x.

Using that, we have that (1-1/x)^2=1, so 1+1/x^2 -2/x=1, and x^2+1-2x=x^2, therefore 2x=1, and x=1/2.

Operating that we have that y=1-2, so y=-1, and x-y=1/2-(-1)=3/2 or 1.5

x y = x/ y

=> y^2 = 1

=> y = +/- 1

With y = 1, x = x = x + 1 cannot be valid x.

With y = -1, - x = - x = x - 1 => x = 0.5

x - y = 0.5 - (-1) = 1.5

Given, xy= $\frac{x}{y}$ =x+y

Clearly, by Law of Transitivity, we have

xy=x+y

Which will give us,

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

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

Given that xy= $\frac{x}{y}$

We get,

y= $\frac{x}{y}$ * $\frac{1}{x}$

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

We substitute this temporary value of y to x= $\frac{x}{y}$ +1

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

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

x=xy+1

Notice that from xy=x+y, we get x=xy-y

With x=xy-y and x=xy+1, by the laws of transitivity, we have

xy-y=xy+1

-y=1

y= -1

We can now use the value of y to get the value of x.

Given, xy= $\frac{x}{y}$ =x+y

x(-1)= $\frac{x}{-1}$ =x+(-1)

-x=-x=x-1

or simply,

-x=x-1

-2x=-1

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

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

Since y= -1 and x= $\frac{1}{2}$ , therefore

x-y= $\frac{3}{2}$ or 1.5

The solution is similar to Just a simple question #2 solution