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I am new to working with modular equations and am trying to find all the solutions to the equation $ 6x \equiv 9\pmod {15} $, here's what I have so far:

$6x \equiv 9 \pmod{15}$ means $6x-9=15m$ for some $m\in \Bbb Z$. Therefore $$6x=15m+9$$ $$2x=5m+3,$$ which shows that $2 \mid (5m+3)$. Since $2 \mid (5m+3)$, we know that $2 \nmid 5m$ (otherwise $2$ would divide $3$, but $2 \nmid 3$). Therefore $2 \nmid m$ (otherwise $2$ would divide $5m$, but $2 \nmid 5m$). Therefore $m=2k+1$ for $k \in \Bbb Z$. Thus we have $$2x=5m+3$$ $$2x=5(2k+1)+3$$ $$2x=10k+5+3$$ $$2x=10k+8$$ and therefore,

$$x=5k+4,$$ so that $x-4=5k$, and hence $x \equiv 4\pmod 5$. Therefore $x \in [4]$ (in modulo 5 arithmetic). I know that $ 5 \mid 15$ and that the congruence classes should be $x \in[4], x\in [9],$ and $x \in [14]$ (modulo 15). The problem is I'm not sure how to get back to mod 15 from mod 5/ what is the process of justification.

Mu_Coffee
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    Your solution looks great. You can plug in 4, 9, and 14 into 6x and see that these are all equal to 9 mod 15. – ndhanson3 Mar 06 '21 at 08:29
  • You're done. All these will be solutions. – Berci Mar 06 '21 at 08:31
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    You have $x=5k+4$ . Now let $k=3j+r$, for any $j$ and $0\le r<3$ – PM 2Ring Mar 06 '21 at 08:46
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    As in the theorem proved in the linked dupe, by the $\rm\color{#c00}{mod\ distributive\ law}$ we have

    $$\begin{align} \bmod 15!:,\ x&\equiv 4 + 5k\ &\equiv 4 + \color{#c00}{5k \bmod 15}\ &\equiv 4 + \color{#c00}{5(k \bmod 3)}\ &\equiv 4 + 5{0,,1,,2} \ &\equiv\ {4,,9,,14} \end{align}$$

     – Bill Dubuque Mar 06 '21 at 09:38

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