Can this system of congruences be solved?

Question

\begin{equation} \left\{\begin{array}{@{}l@{}} 2x\equiv7\mod9 \\ 5x\equiv2\mod6 \end{array}\right.\,. \end{equation} Can this system of congruences be solved? I notice that $(9,6) = 3 \ne 1$ so I can't apply the Chinese theorem of remainders, but this doesn't imply that it can't be solved, so I thought to rewrite the second equation in two different equations, like this:

\begin{equation} \left\{\begin{array}{@{}l@{}} 2x\equiv7\mod9 \\ x\equiv0\mod2 \\ 5x\equiv2\mod3 \end{array}\right.\,. \end{equation}

But the problem still here: $(9,3) = 3 \ne 1$, so it can be solved or is it impossible?

If you want a more general approach: work with the gcd. If a solution existed, we'd have $2x\equiv 7\pmod 3$ AND $5x\equiv 2\pmod 3$. Is that possible? Now there are only three values to test... — lulu, Feb 28 '21 at 20:17
Right. Since $3$ divides $9$ we get that $2x\equiv 7 \pmod 9\implies 2x\equiv 7 \pmod 3$. Similarly for the $\pmod 6$ congruence. — lulu, Feb 28 '21 at 20:25
By the theorem in the linked dupe it is solvable $\iff 2\cdot 2\equiv 7\cdot 5\pmod{3},,$ so it's unsolvable. — Bill Dubuque, Mar 01 '21 at 08:42

score 1 · Answer 1 · answered Feb 28 '21 at 20:30

Hint:

The system of congruences $\begin{cases} x\equiv \alpha\mod a,\\ x\equiv \beta \mod b, \end{cases}$ has a solution if & only if $\alpha\equiv \beta\mod\gcd(a,b)$.

So, solve each congruence independently from the other, and check whether the condition is satisfied.

score 1 · Accepted Answer · answered Feb 28 '21 at 20:36

1

$9$ and $6$ are both multiples of $3$.

$2x\equiv7\pmod9\implies 2x\equiv1\pmod3$

$5x\equiv2\pmod6\implies 2x\equiv2\pmod3$

$1\equiv2\pmod3$ is a contradiction, so the system cannot be solved.

answered Feb 28 '21 at 20:36

J. W. Tanner

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Can this system of congruences be solved?

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