XY matters

$x^{2}-y^{2}=4-2xy$

$(x+y)(x-y)=2(2-xy)$

$2(x-y)=2(2-xy)$

$x-y=2-xy$

subtitute 2 to $x+y$

$x-y=x+y+xy$

$-2y=-xy$

So, we get $y=0 x=2$

So, $x-y=2$ .

$x^2-y^2=4-2xy$

$x^2+y^2=4-2xy$

$\rightarrow 2y^2=0 \rightarrow y=0 \rightarrow x=2 \rightarrow x-y = 2$

From the first equation we have that $(x + y)(x - y) = 2(2 - xy).$

Then with $x + y = 2$ we see that $x - y = 2 - xy,$ and so

$(x + y) - (x - y) = 2 - (2 - xy) \Longrightarrow 2y = xy \Longrightarrow y(x - 2) = 0.$

So either $y = 0,$ which would in turn imply that $x = 2 - y = 2,$ or

$x - 2 = 0 \Longrightarrow x = 2,$ which would in turn imply that $y = 2 - x = 0.$

Either way we find that $x = 2, y = 0,$ and so $x - y = 2 - 0 = \boxed{2}.$

Put y=2-x in the 1st eq. Which gives X^2- 4x+4=0 Then x=2 Y=0 and x-y=2 Hence solved.

$x+y=2 => x^{2}+y^{2}+2xy=4$ summing both ecuations $4+x^{2}-y^{2}=4-2xy+x^{2}+y^{2}+2xy$ $=> y=0 =>x+y=x-y$

$(x+y)^2 = 4$

$\Leftrightarrow x^2 - y^2 = (x^2 + 2xy + y^2) - 2xy$

$\Leftrightarrow x^2 - y^2 = x^2 + y^2$

$\Leftrightarrow y=0$

$\Leftrightarrow x=2$

$\Leftrightarrow x-y=2$

$x^2-y^2=4-2xy$

SO: $x^2+2xy-y^2=4$ (1)

$x+y = 2$

SO: $(x+y)^2=x^2+2xy+y^2=4$ (2)

equating (1) to (2): $y^2=-y^2$

which is only true for $y=0$

$x+y = 2$ SO $x=2$

$x^2-y^2=4-2xy...(i)$

$x+y=2...(ii)$

Squaring both sides in $(ii),$

$x^2+2xy+y^2=4$

Substituting in (i),

$x^2-y^2=(x^2+2xy+y^2)-2xy$

$x^2-y^2=x^2+y^2$

$y^2=0$

$y=0$

From $(ii), x=2$

$x-y=2-0=2$

Hence, the answer is $\boxed{2}$

$x+y=2\\ { x }^{ 2 }-{ y }^{ 2 }=4-2xy\Rightarrow 2xy=4+{ y }^{ 2 }-{ x }^{ 2 }\\ { (x+y) }^{ 2 }={ x }^{ 2 }+2xy+{ y }^{ 2 }\\ { (x+y) }^{ 2 }={ x }^{ 2 }+(4+{ y }^{ 2 }-{ x }^{ 2 })+{ y }^{ 2 }\\ (x+y)^{ 2 }=2{ y }^{ 2 }+4\\ 4=2{ y }^{ 2 }+4\Rightarrow y=0\\ x+y=2\Rightarrow x=2\\ x-y=2$

The answer is 2.

Given: $x^2-y^2=4-2xy$ ....[1]

$x+y=2$ ...............................[2]

On equation [1], add $2xy$ to both sides of the equation. You'll get:

$x^2+2xy-y^2=4$ ...............[3]

Now square both sides of equation [2], and you'll get:

$x^2+2xy+y^2=4$ ...............[4]

Subtract equation [3] from equation [4], the result will become:

$2y^2=0$

$y^2=0$

$y=0$ ......................................[5]

Substituting [5] into [2],

$x+0=2$

$x=2$

Now, since we got already the values of $x$ and $y$ , we can get the value of $x-y$ .

Substituting the values we got for $x$ and $y$ , $x-y=2-0=2$ .

Thus, the answer is 2 .

Given the system of equation: $x^{2} - y^{2} = 4 - 2xy$ $(Eqn. 1)$ and $x^{2} + 2xy + y^{2} = 4$ $(Eqn. 2)$ Reduce the system makes up as: $2x^{2} + 2xy = 8 - 2xy$ Factor 2 both sides yields to: $x^{2} + xy = 4 - xy \Longrightarrow x^{2} + 2xy = 4$ Hence, let $x^{2} = 4 - 2xy$ . So, $x^{2} - y^{2} = x^{2}$ . Reduce $x^{2}$ both sides yields to $y^{2} = 0$ . Since $x^{2} = 4, x - y = 2$ . So, therefore, $x^{2} - y^{2} = (x-y) (x+y) = \boxed{4}.$

x^2-2xy+y^2=4<br> x^2+2xy-y^2=4<br> <br> 2x^2=8<br> <br> <br> x=2<br> <br> y=0<br>

my approach was............

W.K.T: $x^2-y^2+2xiy=(x+iy)^2$

applying it we get: $(x+iy)^2=4$ or x+iy=2 since there is no imaginary part y=0, x=2, x-y=2-0=2