From the first system of equations:

1) x^2 + 2y = 6
2) y = x - 1

From equation 2, we can substitute y in equation 1:

x^2 + 2(x - 1) = 6
x^2 + 2x - 2 = 6
x^2 + 2x - 2 - 6 = 0
x^2 + 2x - 8 = 0
(x + 4)(x - 2) = 0

So, x = -4 or x = 2.

If x = -4, then y = -4 - 1 = -5.
If x = 2, then y = 2 - 1 = 1.

Therefore, the solution to the first system of equations is (-4, -5) and (2, 1).

From the second system of equations:

1) x^2 - y^2 = 24
2) x - 2y = 7

From equation 2, we can express x in terms of y:

x = 7 + 2y

Substitute x in equation 1:

(7 + 2y)^2 - y^2 = 24
49 + 28y + 4y^2 - y^2 = 24
4y^2 + 28y + 49 - y^2 - 24 = 0
3y^2 + 28y + 25 = 0
(3y + 5)(y + 5) = 0

So, y = -5 or y = -5/3.

If y = -5, then x = 7 + 2(-5) = -3.

If y = -5/3, then x = 7 + 2(-5/3) = 7 - 10/3 = 11/3.

Therefore, the solution to the second system of equations is (-3, -5) and (11/3, -5/3).