Solve this systems of equations please! :)

I have been working on this homework problem for a half an hour and can't get it! Can you?
(If there is a 2 after the variable, it means the variable is squared.)

2x² + 8y² + 8x - 48y + 30=0
2x² - 8y² = -48y + 90

Thank you so much!

Divide by 2 and re-arrange.
x² + 4x + 4y² - 24y + 15 = 0
x² - 4y² + 24y - 45 = 0
Re-order in preparation for completing squares.
x² + 4x + 4 + 4y² - 24y + 36 - 36 - 4 + 15 = 0
x² - 4y² + 24y - 36 + 36 - 45 = 0
Complete squares.
(x + 2)² + 4(y - 3)² = 25
x² - 4(y - 3)² = 9
Eliminate 4(y - 3)² between these two equations.
(x + 2)² + x² - 9 = 25
x² + 4x + 4 + x² - 9 = 25
2x² + 4x - 30 = 0
x² + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = -5 or x = 3
Substitute into x² - 4(y - 3)² = 9
25 - 4(y - 3)² = 9 leads to (y-3)² = 4 so y = 5 or y = 1
9 -4(y - 3)² = 9 leads to y = 3
Solutions: (-5, 5) and (-5, 1) and (3, 3)

Start by dividing through both equations by 2 to get smaller numbers to work with:

(2x² + 8y² + 8x - 48y + 30 = 0) / 2 = x^2 + 4y^2 + 4x - 24y + 15 = 0

(2x² - 8y² = -48y + 90) / 2 = x^2 - 4y^2 = -24y + 45

Put the second equation into the same form as the first by adding 24y to both sides and subtracting 45 from both sides:

x^2 - 4y^2 + 24y - 45 = 0


Since both equations have y^2 and y terms with the same (but opposite sign) coefficients, you can add the equations to eliminate the y variable:

x^2 + 4y^2 + 4x - 24y + 15 = 0
+(x^2 - 4y^2 + 24y - 45 = 0)
--------------------------------------…
2x^2 + 4x - 30 = 0

Divide through by 2:

x^2 + 2x - 15 = 0

factor to:

(x + 5)(x - 3) = 0

x = -5 and x = 3


Substitute those values for x in either original equation to solve for the two values of y.