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My question is as following:

How do I solve:

$$3x+2y \equiv 2 \ \ (mod 5) $$

$$x+y \equiv 3 \ \ (mod 4)$$

My thought is to use Chinese remainder theorem to first find a solution such that:

$$z \equiv2 \ \ (mod5)$$

$$z \equiv 3 \ \ (mod 4)$$

Applying CRT, I got

$$z \equiv 7 \ \ (mod 20)$$

And for each possible value of $z$, we have a unique solution for $x$ and $y$, but I am wondering if it's the suitable way of solving these kind of problem.

aawangas
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  • What is $z$? $3x+2y$ or $x+y$? You seem to think it is both. – Anne Bauval Dec 17 '23 at 08:52
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    I think $z$ can be a set of value such that $z mod 20 = 7$, so that taking out any of them will give us a linear system of $x$ and $y$. For example: $3x+2y = 7, x+y = 27$ – aawangas Dec 17 '23 at 09:01
  • $3x+2y \equiv 2\pmod5$, not necessarily $\pmod{20}$. Similarly, $x+y \equiv 3\pmod 4$, not necessarily $\pmod{20}$. – Anne Bauval Dec 17 '23 at 09:08
  • I see. So is there anyway I can find other solutions? – aawangas Dec 17 '23 at 09:45
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    Choose for instance any $y\in\Bbb Z$, and then solve$$3x\equiv2-2y\pmod5,\quad x\equiv3-y\pmod4$$ i.e. $$x\equiv4-4y\pmod5,\quad x\equiv3-y\pmod4.$$ (I found $x\equiv-1-9y\pmod{20}$). – Anne Bauval Dec 17 '23 at 11:54

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