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I wish to solve the following system of congruences using the Chinese remainder theorem: $$13X \equiv 3 (\text{mod} 15), \\ 2X \equiv 6 (\text{mod} 10).$$ I have reduced this system to $$X \equiv 6 (\text{mod} 15), \\ X \equiv 3(\text{mod} 5),$$ by multiplying the first by the multiplicative inverse of $13$ and dividing the second through by $2$. However, the moduli are still not relatively prime, so I am unsure how to solve this system.

V. Elizabeth
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    $\pmod 5$ the first one gives us $x\equiv 1 \pmod 5$ but the second gives us $2x\equiv 1 \pmod 5$, so not compatible. – lulu Sep 28 '23 at 18:44
  • @lulu How does the first give $x \equiv_{5} 1$? – V. Elizabeth Sep 28 '23 at 18:55
  • @V.Elizabeth What do you mean? $13\equiv 3\pmod 5$ and $3x\equiv 3\pmod 5$ is the same as $x\equiv 1 \pmod 5$. – lulu Sep 28 '23 at 19:03
  • They're inconsistent mod the moduli gcd $5!:\ \color{#c00}{x\equiv 1},$ vs. $,\color{#0a0}{x\equiv 3}, $ (i.e. $,\color{#c00}{x\equiv3/13}\not\equiv \color{#0a0}{6/2\equiv x}),,$ so they are unsolvable. See General Easy CRT in the linked dupe for CRT for noncoprime moduli. – Bill Dubuque Sep 28 '23 at 19:09

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