In this case we have two equations:

(1) -7x - 2y = 11

(2) 4x + y = - 6 ==> (2a) y = - 6 - 4x

We can substitute (2a) to (1)

=> -7x - 2 (-6 - 4x) = 11

=> -7x + 12 + 8x = 11

=> -7x + 8x = 11 - 12

=> x = -1

Then we substitute x = -1 to either (1) or (2). Here, I substitute x = -1 to (1)

=> -7 (-1) - 2y = 11

=> -2y = 11 - 7

=> -2y = 4

=> y = 4/-2 = -2

Thus, (x,y) => (-1, -2)

Ellimination:

First I will times 2 the (2)' equation, so

=> 2* (4x + y = -6)

=> 8x + 2y = -12 .....(2b)

Now we will elliminate 2y by using (1) & (2b)

-7x - 2y = 11

8x + 2y = -12 +

______________

x = -1

Then, I substitute x = -1 to either (1) & (2). Here I subtitute it to (2), and I get:

=> 4 (-1) + y = -6

=> -4 + y = -6

=> y = -6 + 4

=> y = -2

Thus, (x,y) => (-1, -2)

Both techniques come up with the same result.

In this case, the substitution method seems more easier than elimination.

We can re-write the 2nd equation keeping y to the left side:

y = -4x - 6

Now, we'll substitute the expression of y in the 1st equation:

-7x - 2(-4x-6) = 11

We'll remove the brackets:

-7x + 8x + 12 = 11

We'll combine like terms:

x = 11 - 12

x = -1

y = -4*(-1) - 6

y = 4 - 6

y = -2

The solution of the system is represented by the pair (-1;-2).

To apply the elimination method, we'll have to multiply the 2nd equation by 2.

8x+2y=-12

Now, we'll add this equation to the 1st one:

-7x-2y+8x+2y=11-12

We'll eliminate like terms:

x=-1

We'll substitute x in the 1st equation:

7 - 2y = 11

-2y = 11-7

-2y = 4

y = -2

We notice that in both cases, we've get the same result.

**Therefore, the solution of the system is represented by the pair (-1;-2).**