We can rewrite them as:
$\displaystyle 2x\equiv 1\equiv 8 (mod \;\ 7)\Rightarrow \boxed{x\equiv 4(mod \;\ 7)}$
$\displaystyle 5x\equiv 2\equiv 13\equiv 24\equiv 35(mod 11)\Rightarrow \boxed{x\equiv 7(mod \;\ 11)}$
From these we can write:
$\displaystyle 7a+4=11b+7$...[1]
Appyling modulo 7 to $\displaystyle 7a+4=11b+7$ and we get:
$\displaystyle b\equiv 1(mod \;\ 7)$
$\displaystyle b=7c+1$
Sub into the right side of [1]:
$\displaystyle 11(7c+1)+7=77c+18$
The solutions to the system are:
$\displaystyle x\equiv 18(mod \;\ 77)$
Which can be written as $\displaystyle \boxed{x=77t+18}$
Let $\displaystyle t=0,1,2,......$ and they will satisfy the system of linear congruences.