The question is: Find $x, y ∈ \mathbb Z_{11}$ that satisfy the simultaneous system of linear equations
$[5]_{11} x +[2] _{11} y = [6] _{11}$
$[4]_{11} x + y = [2] _{11}$
So I rewrote this as
$5x + 2y = [6] _{11}$ ,
$4x + y = [2] _{11} $, which is equivalent to
$5x + 2y = 6$ (mod $11) (1)$
$4x + y = 2$ (mod $11) (2)$
$2(2) - (1)$ gives $3x = -2$ (mod $11) ⇒ 3x = 11k - 2$, for some integer $k$.
Then $(1) + (2)$ gives $9x + 3y = 8 ($mod $11) ⇒ 3(3x) + 3y = 8 ($mod $11) ⇒ 33k - 6 + 3y = 8$ (mod $11) ⇒ 3y + 33k = 14$ (mod $11) = 3 ($mod $11) ⇒ y + 11k = 1 ($mod $11) ⇒ y = 11m + 1 ⇒ y = [1] _{11} $
Then $(1)$ gives $5x + 2 = 6 ($mod $11)$, and, by noticing that $17 = 5(3) + 2$ and $17 = 6 ($mod $11)$, I got $x = [3] _{11} $.
So, first of all, is my rewriting the congruence class equations as modular equations correct? And, if so, is my solution right, and are there any more solutions?