several equations , determine (x,y)?

add then together to get 2 x² = 2x ===> x = 0 or 1....if x = 0 then y² = y ===> y = 0 or 1....if x = 1 then  y² = - y ===> y = 0 or - 1 ; if x = 0 then y² = y ===> y = 0 or 1......(1,0) , ( 1 , - 1 ) and (0, 0 ) , ( (0,1 )

(1) : x² + y² = x - 2xy + y

(2) : x² - y² = x + 2xy - y

You calculate (1) + (2)

(x² + y²) + (x² - y²) = (x - 2xy + y) + (x + 2xy - y)

x² + y² + x² - y² = x - 2xy + y + x + 2xy - y

2x² = 2x

x² = x

x² - x = 0

x.(x - 1) = 0

First case: x = 0 → the system of the 2 equation becomes:

(1) : x² + y² = x - 2xy + y → where: x = 0 → y² = y

(2) : x² - y² = x + 2xy - y → where: x = 0 → - y² = - y → y² = y

You obtain only one equation:

y² = y

y² - y = 0

y.(y - 1) = 0

First possibility: y = 0

Second possibility: (y - 1) = 0 → y - 1 = 0 → y = 1

2 solutions (x ; y) → (0 ; 0)   (0 ; 1)

Second case: (x - 1) = 0 → x - 1 = 0 → x = 1 → the system of the 2 equation becomes:

(1) : x² + y² = x - 2xy + y → where: x = 1 → 1 + y² = 1 - 2y + y → y² = - y

(2) : x² - y² = x + 2xy - y → where: x = 1 → 1 - y² = 1 + 2y - y → - y² = y → y² = - y

You obtain only one equation:

y² = - y

y² + y = 0

y.(y + 1) = 0

First possibility: y = 0

Second possibility: (y + 1) = 0 → y + 1 = 0 → y = - 1

2 solutions (x ; y) → (1 ; 0)   (1 ; - 1)

Resume the solutions (x ; y)

(0 ; 0)   (0 ; 1)   (1 ; 0)   (1 ; - 1)

Rewrite the first eqs as x² + 2xy + y² = x+y --> (x+y)² = x+y.

Two cases: x+y=0 thus x=-y, OR x+y = 1 thus x = 1-y.

Replace each of these into the other eqs. Your solutions will follow.

Done!

This is a system of two equations and two unknowns:

x² + y² = x - 2xy + y and x² - y² = x + 2xy - y

Let's solve the first equation for y in terms of x.  First, I'll move everything to one side then group the like-terms together to make a quadratic.  Any term with only "x" in it will be considered a constant for now:

x² + y² - x + 2xy - y = 0

y² + 2xy - y + x² - x = 0

y² + (2x - 1)y + (x² - x) = 0

Now we have a quadratic where:

a = 1

b = 2x - 1

c = x² - x

Quadratic equation:

y = [ -b ± √(b² - 4ac)] / (2a)

y = [ -(2x - 1) ± √((2x - 1)² - 4(1)(x² - x))] / (2 * 1)

y = [ -2x + 1 ± √(4x² - 4x + 1 - 4(x² - x))] / 2

y = [ -2x + 1 ± √(4x² - 4x + 1 - 4x² + 4x)] / 2

The x and x² terms cancel out under the square root:

y = [ -2x + 1 ± √(1)] / 2

y = (-2x + 1 ± 1) / 2

y = (-2x + 1 - 1) / 2 and y = (-2x + 1 + 1) / 2

y = -2x / 2 and y = (-2x + 2) / 2

y = -x and y = -x + 1

Now we have two expressions for y in terms of x.  If we substitute them into the other equation we will have two additional quadratics to solve:

x² - y² = x + 2xy - y

x² - (-x)² = x + 2x(-x) - (-x) and x² - (-x + 1)² = x + 2x(-x + 1) - (-x + 1)

x² - x² = x - 2x² + x and x² - (x² - 2x + 1) = x + (-2)x² + 2x + x - 1

0 = -2x² + 2x and x² - x² + 2x - 1 = -2x² + 4x - 1

0 = -x² + x and 2x - 1 = -2x² + 4x - 1

0 = -x² + x and 0 = -2x² + 2x

0 = -x² + x and 0 = -x² + x

Now that both equations are the same, we can solve one to get solutions for both:

0 = -x² + x

0 = (-x + 1)x

-x + 1 = 0 and x = 0

1 = x and x = 0

Now that we have two values for x we'll put them into both expressions for y to see what we end up with:

y = -x and y = -x + 1

y = -0 and y = -0 + 1 and y = -1 and y = -1 + 1

y = 0 and y = 0 + 1 and y = -1 and y = 0

y = 0 and y = 1 and y = -1 and y = 0

When x = 0 we have y = 0 and 1.  That gives us the two points (0, 0) and (0, 1).

When x = 1 we have y = -1 and 0.  That gives us the two points (1, -1) and (-1, 0).

All four points are a solution to the system of equations.