SOLUTION: (x+1)^2-(y-1)^2=20, x^2-(y+2)^2= 24

You are given the system of two equations in two unknowns; each equation is of degree 2:

(x+1)^2-(y-1)^2 = 20,          (1)
x^2    -(y+2)^2 = 24.          (2)


Your first step is to expand:

x^2 + 2x + 1 - y^2 + 2y - 1 = 20,
x^2          - y^2 - 4y - 4 = 24.


Simplify:

x^2 + 2x - y^2 + 2y = 20,      (3)
x^2      - y^2 - 4y = 28.      (4)


Subtract (4) from (3) (both sides):

     2x + 6y        =  -8,     (5)
x^2      - y^2 - 4y =  28.     (6)


Now you have one equation of degree 1 (linear equation (5)) and one equation of degree 2.

So, you reduced the degree,  and it was your first major step.


Now, multiply eq(6) by 4 (both sides)

     2x + 6y        =   -8,    (7)
4x^2    - 4y^2 - 16y = 112.    (8)


From (7), express  2x = -8- 6y;  ====>  4x^2 = (-8 - 6y)^2 = (8 + 6y)^2 and substitute this expression for 4x^2 into (8). 
You will get

(8 + 6y)^2 - 4y^2 - 16y = 112   (9)    instead of (8).


Thus you reduced the system to the single quadratic equation for y, and it was our second major step.


Now it is simple arithmetic to solve it.


When you get  y,  calculate  x. 

    First major step is to reduce the given system of two eqs of degree 2 to 
    one equation of degree 2 and one eqn of degree 1 (linear eqn).

    It is "the degree reducing step".


    Second major step is reducing the simplified system to a single quadratic equation.
    It is achieved using substitution from the linear equation.


    When these major steps are completed, the rest is arithmetic.