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I have the following equations:

$$8\equiv-(3a+7b)\mod11$$ and $$9\equiv-(5a+4b)\mod11$$

How do I solve them? I've tried to subtract one equation from the other, but I end up with the wrong answer.

It is supposed to be: $a=2$ and $b=9$.

Micah Windsor
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Benjamin H
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    The method is the same as it always is for solving a system of linear equations. If you show your calculation, perhaps we can spot where you went astray. – lulu Apr 10 '20 at 17:47
  • Hint $\bmod 11!:\ 7\equiv -4,$ so adding them eliminates $,b,,$ yielding $,6\equiv 3a\ \ \ $ – Bill Dubuque Apr 10 '20 at 19:12
  • Worth emphasis: the (Gaussian) elimination method works the same $!\bmod 11,$ because the modulus is prime so we are working over a field, so $,a\not\equiv 0,\Rightarrow, a^{-1}$ exists. This fails for nonprime moduli so we need to modify our methods in this case. – Bill Dubuque Apr 10 '20 at 19:31

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By rearranging the first congruence equation $8 \equiv -3a -7b \pmod{11}$, we have $4b \equiv 8 + 3a \pmod{11}$. Since $\gcd(4, 11) = 1$, there is an inverse of $4$, which is $3$. We thus have \begin{equation} b \equiv 24 + 9a \equiv 2 - 2a \pmod{11} \end{equation} Plugging this into the other congruence equation $9 \equiv -5a - 4b \pmod{11}$. We have \begin{equation} 9 \equiv -5a - 4(2 - 2a) \equiv 3a - 8 \pmod{11} \end{equation} which implies that $3a \equiv 17 \equiv 6 \pmod {11}$. Since $\gcd(3, 11) = 1$, we can divide by $3$. So, we have $a \equiv 2 \pmod{11}$. Using the same way, you can solve for $b$.

saru
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