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I don't know where to start and how to go forth when solving system of equations in for example $\mathbb{Z}_{11}$. I have 2 different systems I want help with with a walkthrough to understand what is going on.

System 1:

\begin{align}4x + 7y &= 3 \pmod {11}\\ 8x + 5y& = 9 \pmod {11}\end{align}

System 2:

\begin{align}3x + y + z &= 1 \pmod {11}\\ 5x + y - z &= 2 \pmod{11}\\ 2x - y - z &= 1 \pmod {11}\end{align}

lioness99a
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user633788
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  • In the same way if you work over the real numbers. In ${\Bbb Z}_{11}$ each element has an additive inverse and each nonzero element has a multiplicative inverse. – Wuestenfux Jan 11 '19 at 14:23
  • Note: as $11$ is prime, your ring is a field. You can perform division without shame – Damien Jan 11 '19 at 14:25

2 Answers2

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What's peculiar about $\mathbb{Z}_{11}$? You do it as you would do it over any other field. Consider the first system:$$\left\{\begin{array}{l}4x+7y=3\\8x+5y=9.\end{array}\right.$$First step: you divide the first equation by $4$. Since we're working over $\mathbb{Z}_{11}$ here, that's the same thing as multiplying by $3$, thereby getting$$\left\{\begin{array}{l}x+10y=9\\8x+5y=9.\end{array}\right.$$Now, you subtract from the second equation the first one times $8$ (which is the coefficient of $x$ from the second equation):$$\left\{\begin{array}{l}x+10y=9\\2y=3.\end{array}\right.$$Now you know that $y=7$ and so $x=\cdots$

Can you solve the other system now?

José Carlos Santos
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To expand slightly on a previous answer, you have

\begin{align}4x+7y&=3\pmod {11}\tag{1}\\ 8x+5y&=9\pmod {11}\tag{2}\end{align}

As with regular simultaneous equations, we want to multiply $(1)$ by $2$ so we have an $8x$ in both equations:

\begin{align}2\times (4x+7y)&=2\times3\\ 8x+14y&=6\end{align}

However, this gives us $14y$ which we cannot have as we are working in $\mathbb{Z}_{11}$. However, $$14\pmod{11}=3$$ so instead we can write $$8x+3y=6\tag{$(3)=(1)\times 2$}$$

Now we can do $(2)-(3)$ to get

\begin{align}(8x+5y)-(8x+3y)&=9-6\\ 8x-8x+5y-3y&=3\\ 2y&=3\end{align}

We cannot simply say that $y=\frac 32$ as we cannot have fractions in $\mathbb {Z}_{11}$, but we know that, as $11$ is prime, each element will have a multiplicative inverse, and so we can find that $y=7$ as \begin{align}2y&=2\times 7\\ &=14\\ &=3\pmod{11}\end{align}

We can then substitute this value into $(2)$ to get \begin{align}8x+5y&=9\\ 8x+5\times 7&=9\\ 8x+35&=9\\ 8x+2&=9\tag{*}\\ 8x&=7\end{align}

$(*)$ We replace the $35$ with $2$ because $35\pmod{11}=2$

Once again, we know that $x=5$ because \begin{align}8x &= 8\times 5\\ &=40\\ &=7\pmod{11}\end{align}

Therefore, the solution to our simultaneous equations is $x=5, y=7$. We can check this in $(1)$: \begin{align}4x+7y&=4\times 5+7\times 7\\ &= 20 + 49\\ &= 69\\ &= 3\pmod {11}\end{align}

lioness99a
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