x^2+y^2-2xy=21

x^2+2xy-8y^2=0
find x and y
help me plz

x^2+2xy-8y^2=0

(x+4y)(x-2y) = 0
so two routes to follow

x = 2 y and x = -4y

now the first one with x = 2 y

4 y^2 + y^2 - 4 y^2 = 21
y = +21 or - 21
then x = 42 or -42
so
(42,21) (-42,-21)

now the first one with x = -4y
I will leave that for you

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To find the values of x and y that satisfy the given equations, we can use a method called substitution. Here's how you can solve the system of equations step by step:

Step 1: Solve the second equation for x in terms of y.
x^2 + 2xy - 8y^2 = 0
x^2 = 8y^2 - 2xy
x^2 = 2y(4y - x)
x = ±√(2y(4y - x))

Step 2: Substitute the value of x from Step 1 into the first equation.
(x^2 + y^2 - 2xy = 21)
((±√(2y(4y - x)))^2 + y^2 - 2(±√(2y(4y - x)))y = 21)

Simplify the first equation:
2y(4y - x) + y^2 - 2(±√(2y(4y - x)))y = 21

Step 3: Simplify the equation obtained in Step 2 and solve for y.
8y^2 - 2xy + y^2 - (±√(2y(4y - x)))2y = 21
9y^2 - 4xy - (±2√(2y(4y - x)))y = 21

Now, we have a quadratic equation in terms of y. Let's simplify it further:

9y^2 - 4xy - 2y√(2y(4y - x)) = 21

Step 4: Solve the quadratic equation from Step 3 using any method you prefer (factoring, quadratic formula, completing the square, etc.). Once you find the values of y, substitute them back into the expression for x obtained in Step 1 to find the corresponding values of x.