Can you solve this equition: x^2 - y^2 + 1 = 5xy - 5y. Solve the x and y... Thanks in advance.?

Answer 1

x^2-y^2+1=5y(x-1)
putting x=1
1-y^2+1=0
y^2=2
y=+/-rt2
similarly byascribing values to x we can getthe corresponding values of y

aliter
x^2-5xy-(y^2-5y-1)=0
x=[5y+/-rt{25y^2+4(y^2-5y-1)}]/2
=[5y+/-rt{29y^2-20y-4}]/2
now for differentvalues of y we can get the correspondingvalues of x

Answer 2

This is a simple cubic: x^3-yx+y=0, which has the following three solutions: x1= -((2 3^(1/3) y + 2^(1/3) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^(2/3))/( 6^(2/3) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^(1/3))) x2= ( 2 (3 i + Sqrt[3]) y + 2^(1/3) 3^( 1/6) (1 - i Sqrt[3]) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^(2/3))/( 2 2^(2/3) 3^(5/6) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^( 1/3)) x3= ( 2 (-3 i + Sqrt[3]) y + 2^(1/3) 3^( 1/6) (1 + i Sqrt[3]) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^(2/3))/( 2 2^(2/3) 3^(5/6) (9 y + Sqrt[3] Sqrt[(27 - 4 y) y^2])^(1/3))

Answer 3

Set 5xy-5y =0
This gives x =1
Set x^2-y^2+1= 0
Put x=1 in above equation, getting
1 -y^2 +1 = 0 so ,
-y^2=-2
y= +/- sqrt(2)

This is a solution to your problem

Answer 4

ummm..... 5x^3y+5y^3=1 no?