If you're supposed to solve the congruences one at a time:
1: [tex]2x\equiv5\pmod7[/tex]
First find the inverse of 5 modulo 7. Note that [tex]5\cdot3\equiv15\equiv1\pmod7[/tex], so [tex]5^{-1}\equiv3\pmod7[/tex].
[tex]3\cdot2x\equiv6x\equiv1\pmod7[/tex]
Then find the inverse of 6 modulo 7. Note that [tex]6\cdot6\equiv36\equiv1\pmod7[/tex], so 6 is its own inverse, and we have
[tex]6\cdot6x\equiv x\equiv6\pmod7[/tex]
so the first congruence has solution [tex]x=6+7n[/tex] where [tex]n\in\mathbb Z[/tex].
2: [tex]4x\equiv2\pmod5[/tex]
We have [tex]3\cdot2\equiv6\equiv1\pmod5[/tex], so [tex]2^{-1}\equiv3\pmod5[/tex]. Then
[tex]3\cdot4x\equiv12x\equiv2x\equiv1\pmod5[/tex]
and
[tex]3\cdot2x\equiv x\equiv3\pmod5[/tex]
so the solution is [tex]x=3+5n[/tex] for [tex]n\in\mathbb Z[/tex].
3: [tex]3x\equiv9\pmod{11}[/tex]
[tex]9\cdot5\equiv45\equiv1\pmod{11}[/tex], so [tex]9^{-1}\equiv5\pmod{11}[/tex], and
[tex]5\cdot3x\equiv15x\equiv4x\equiv1\pmod{11}[/tex]
[tex]4\cdot3\equiv12\equiv1\pmod{11}[/tex], so [tex]4^{-1}\equiv3\pmod{11}[/tex] and
[tex]3\cdot4x\equiv x\equiv3\pmod{11}[/tex]
so [tex]x=3+11n[/tex] for [tex]n\in\mathbb Z[/tex].
If they're to be taken simultaneously, use the Chinese remainder theorem.
Combined:
[tex]\begin{cases}x\equiv6\pmod7\\x\equiv3\pmod5\\x\equiv3\pmod{11}\end{cases}[/tex]
Let's start with
[tex]x=5\cdot11+7\cdot11+7\cdot5[/tex]
so that two terms will vanish when considering the remainder of [tex]x[/tex] modulo 7, 5, or 11, respectively.
Taken modulo 7, we have
[tex]x\equiv5\cdot11\equiv55\equiv6\pmod7[/tex]
Taken modulo 5, we have
[tex]x\equiv7\cdot11\equiv77\equiv2\pmod5[/tex]
but we want to end up with 3, so we multiply the second term by the inverse of 2 modulo 5, then by 3. Note that [tex]2\cdot3\equiv1\pmod5[/tex], so our new choice of [tex]x[/tex] is
[tex]x=5\cdot11+7\cdot11\cdot3\cdot3+7\cdot5[/tex]
and taken modulo 5, we end up with a remainder of 3, as desired.
Taken modulo 11, we have
[tex]x\equiv35\equiv2\pmod{11}[/tex]
but we want to end up with 3, so we multiply the third term by the inverse of 2 modulo 11, then by 3. Note that [tex]2\cdot6\equiv1\pmod{11}[/tex], so [tex]x[/tex] becomes
[tex]x=5\cdot11+7\cdot11\cdot3\cdot3+7\cdot5\cdot6\cdot3=1378[/tex]
By the CRT, the least positive solution is
[tex]x\equiv1378\pmod{7\cdot5\cdot11}\implies x\equiv1378\pmod{385}\implies x\equiv223\pmod{385}[/tex]
so the general solution to the system would be [tex]x=223+385n[/tex] for [tex]n\in\mathbb Z[/tex].
