Instead of doing your problem for you, I will do one exactly 
in every detail step-by-step like your problem.  All you have
to do is use the problem below as a model. The problem I will
do is:



One thing we observe about this problem is that it is symmetrical
in x and y.  That is we have the same two equations if we interchange
x and y.  Therefore any time we get a pair of solutions for x, one will
be a solution for x and the other will be a solution for y.

Solve the second for y




 

Substitute in the first equation:



Multiply through by (x+2)², [but we know x can't be -2]



Get binomials in descending order:









Looking at the graph on a graphing calculator it appears that 
it has solutions 2 and 5 but it probably has a pair of conjugate
complex solutions too.  We use synthetic division with x=2

2 | 1 1 -6 -200  400
  |   2  6    0 -400 
    1 3  0 -200    0

5 | 1  3   0 -200
  |    5  40  200    
    1  8  40    0

So we have factored the polynomial as



So we have solutions x=2, x=5, 

By the symmetry, we know that when x=2, y=5
and when x=5, y=2.

we get the other solutions from
using the quadratic formula, since it won't factor.

















So the other pair of solutions, by symmetry, are

if , then 

and 

if , then 

Now you do your problem exactly step-by step like I did this one.

Edwin