Solve the system of equations 3x^2-3y^2=27 and 2x-2y=2 .

We'll re-write the first equation factorizing by
3:

3(x^2 - y^2) = 27

We'll
divide by 3:

x^2 - y^2  = 9
(1)

x - y = 1 (2)

We notice
that we can re-write (1) as a product, being a difference of
squares:

(x-y)(x+y) = 9

But
x-y = 1 => x + y = 9 (3)

We'll add (3) to
(2):

x + y + x – y = 9 +
1

We'll eliminate like
terms:

2x = 10

x =
5

We'll substitute x = 5 into
(2):

5 – y = 1

-y = -5 +
1

y = 4

The
solution of the system is represented by the pair {5 ;
4}.