Solve the linear congruence $9x \equiv 21 ~(mod ~ 30)$

 Example: Solve the linear congruence $9x \equiv 21 ~(mod ~ 30)$.

Answer: Since $gcd(9, 30) = 3$ and $3$ divides $21$, by  Theorem, given linear congruence $9x \equiv 21 ~(mod ~30)$ has $3$ mutually incongruent solutions modulo $30$.  We will find first solution by trial and error method. We will consider simple values of $x$ as $1, ~-1,~ 2,~ -2,~ \cdots $ and check which value satisfy given linear congruence $9x \equiv 21 ~(mod~ 30)$. When we take $x = 1$, we observe that $x =1$ is not a solution of $9x \equiv 21 ~(mod ~30)$. Now, we take $x = - 1$, we observe that $x = - 1$ is  a solution of $9x \equiv 21 ~(mod ~30)$. But $-1 \equiv 29 ~(mod ~30) $, therefore $x_{0} = 29$ is a solution of a given  linear congruence $9x \equiv 21 ~(mod ~30)$.  By Theorem, other solutions of given linear congruence $9x \equiv 21 ~(mod ~30)$  are given by $x_{1} = 29 + \frac{30}{3} = 29 + 10 = 39 \equiv 9~ (mod ~ 30), ~ x_2 = 29 + (\frac{30}{3}) ~2 = 29 + 10 \cdot 2 = 29 +20 = 49 \equiv 19 ~(mod ~ 30)$. Therefore $9, ~19$ and $ 29$ are $3$ mutually incongruent solutions modulo $30$ of  given linear congruence $9x \equiv 21 ~(mod ~30)$. 

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