$3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. Given that $3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. I'm thinking $20\equiv9\pmod{11}$, But I am having trouble find a number $3x$ that  is divisible by $3$? Is there a better way of solving this problem.
 A: From the first congruence, by multiplying by $4$, we conclude that $x\equiv 28\equiv 6\pmod{11}$.
From the second congruence, multiplying by $2$, conclude that $-y\equiv 18\equiv 7\pmod{11}$, so $y\equiv -7\equiv 4\pmod{11}$.
Add. We get $x+y\equiv 10\pmod{11}$. 
Another way: Multiply the first congruence by $5$, the second by $3$, and add. We get
$15x+15y\equiv 62\pmod{11}$. This can be rewritten as $4(x+y)\equiv -4\pmod{11}$, which gives $x+y\equiv -1\pmod{11}$. 
A: 3x $\equiv$ 7 (mod 11) and 5y $\equiv$ 9 (mod 11)
3x $\equiv$ 18 (mod 11) by adding 11 to 7, then 5y $\equiv$ 20 (mod 11) by add 11 to 9.
x $\equiv$ 6 (mod 11) by dividing 3 to both sides, then y $\equiv$ 4 (mod 11) by dividing 5 to both sides.
Then, x + y $\equiv$ 10 (mod 11)
A: $3x\equiv7\pmod{11}$ $\implies$ $12x\equiv28\pmod{11}$  $\implies$  $x\equiv6\pmod{11}$
$5y\equiv9\pmod{11}$ $\implies$  $45y\equiv81\pmod{11}$  $\implies$  $y\equiv4\pmod{11}$
Therefore
$x+y\equiv10\pmod{11}$
A: By inspection, $x = 6, y = 4$ is a solution.
For a more general approach, try solving $$3x \equiv 1 \bmod{11}$$ and $$5y \equiv 1 \bmod{11}$$ using Extended GCD, and multiply the first by 7 and the second by 9. To get more solutions, just increase/decrease either $x$ or $y$ by 11. 
(However, since you're just interested in what $x + y$ is mod 11, you can skip the "get more solutions" part; they would all get the same answer.)
A: We have $15x\equiv 35\pmod {11}\equiv2$ and $15y\equiv27\pmod{11}\equiv5$
$\implies 15(x+y)\equiv7\pmod{11}\iff 4(x+y)\equiv7\pmod{11}\equiv40$
$\implies x+y\equiv 10\pmod{11}$ dividing either sides by  $4$ as $(4,11)=1$
