1. ## solve in Z

solve in Z :

2. Note that 8 is a solution to both equations.

11*9=99

8+99 is also a solution. (Why?)
Similarly 8+2*99 is also a solution.

Conclusion: 8+99k is the general solution, where k is any integer.

3. please : more of détailes on resolution

4. Which part of the solution didn't you understand?

5. Originally Posted by alexmahone
Which part of the solution didn't you understand?
resolution génerale

6. The conclusion is valid, the argument less so. If once cites the Chinese Remainder theorem, that argument would be sufficient

Alternatively:

$\displaystyle x \equiv -3 \pmod{11} \Rightarrow x = 11k-3$

Therefore,

$\displaystyle 11k-3 \equiv -1 \pmod{9} \Rightarrow 11k \equiv 2 \pmod{9} \Rightarrow k \equiv 2(11)^{-1} \pmod{9}$
$\displaystyle \Rightarrow k \equiv 1 \pmod{9} \Rightarrow k = 9l + 1 \Rightarrow x = 99l+8$

7. ## Offered resolution

OFFERED RESOLUTION

8. Originally Posted by SimonM
The conclusion is valid, the argument less so. If once cites the Chinese Remainder theorem, that argument would be sufficient

Alternatively:

$\displaystyle x \equiv -3 \pmod{11} \Rightarrow x = 11k-3$

Therefore,

$\displaystyle 11k-3 \equiv -1 \pmod{9} \Rightarrow 11k \equiv 2 \pmod{9} \Rightarrow k \equiv 2(11)^{-1} \pmod{9}$
$\displaystyle \Rightarrow k \equiv 1 \pmod{9} \Rightarrow k = 9l + 1 \Rightarrow x = 99l+8$
HELLO : k differe of k ' thank you