$3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to.

Question

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.

Answer 1

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)

Answer 2

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 $x+y\equiv -1\pmod{11}$.

Answer 3

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.)

Answer 4

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 $(4,11)=1$