Question

x^2+y^2-2xy=21
x^2+2xy-8y^2=0
find x and y
help me plz

Answer 1

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

Answer 2

f gwsertg wreg