<s>[INST] $3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. 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.
 [/INST] 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 $x+y\equiv -1\pmod{11}$. 
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