I need to find all solutions to:
$$4x\equiv3\pmod7$$
I know the solutions are in ${0, 1, 2, 3, 4, 5, 6}$ and I got $x \equiv 6 \pmod7$ so my answer was 6 but I don't know if that's all the answers.
I have the same problem with:
$$3x+1\equiv4\pmod5$$
I got $x\equiv1\pmod5$ so my answer was $1$ since it is in ${0, 1, 2, 3, 4}$.
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For the first problem you need to find the multiplicative inverse of $4$ modulo $7$ and multiply both sides by that. This is the modular arithmetic equivalent of "dividing by 4" to solve for $x$. Here multiplicative inverse of $4$ modulo $7$ means an integer $a$ such that $4 \cdot a \equiv 1 \pmod 7$. You can check by trial and error that $a=2$ works since $4\cdot 2 = 8 \equiv 1 \pmod 7$. Thus, \begin{align*} 4x&\equiv 3 \pmod 7\\ 2\cdot 4 x &\equiv 2 \cdot 3 \pmod 7\\ 8x &\equiv 6 \pmod 7\\ x &\equiv 6 \pmod 7. \end{align*} Since multiplicative inverses are unique modulo prime numbers, this will be the only solution in $\{0,1,\dots,6\}$.
For the second problem you can do something similar, but first move the $1$ to the other side to get $3x \equiv 3 \pmod 5$. From this it's clear that $x=1$ is a solution, but you can also find that by multiplying both sides by $2$ (the multiplicative inverse of $3$ modulo $5$). It's the only solution since again multiplicative inverses are unique modulo $5$.
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Find all solutions to: $$4x\equiv3\pmod7$$
Given
$$4x\equiv 3\pmod 7 \iff 8x\equiv 6\pmod 7 \iff x\equiv 6\pmod 7$$
then as the least residue system modulo $7$ is
$$\{0,1,2,3,4,5,6\}$$
we see that $x\equiv 6\pmod 7$ when $x=6$. Therefore, the only solution is $x\equiv 6\pmod 7$.
Find all solutions to: $$3x+1\equiv4\pmod5$$
Given
$$3x+1\equiv4\pmod5 \iff 3x\equiv3\pmod5 \iff x\equiv 1\pmod5$$
then as the least residue system modulo $5$ is
$$\{0,1,2,3,4\}$$
we see that $x\equiv 1\pmod 5$ when $x=1$. Therefore, the only solution is $x\equiv 1\pmod 5$.
To see that both solutions are unique we apply the following theorem
Theorem: If $\text{g}=\text{gcd}(a,m)=1$, the congruence $ax\equiv b\pmod m$ has exactly one solution modulo $m$.
In the first case, $\text{g}=\text{gcd}(4,7)=1$ and in the second case $\text{g}=\text{gcd}(3,5)=1$. Therefore, both solutions are unique.
In these problems, you can plug in each of the possible values for $x$ (since there are finitely many), and see which ones work. For the first problem, it is indeed true that only $x\equiv 6\pmod{7}$ is a solution, and for the second, it is true that only $x\equiv 1\pmod{5}$ is a solution.
If you know about fields, it turns out that $\Bbb{F}_p = \Bbb{Z}/(p)$ (i.e., the set of numbers you use in modular arithmetic) forms a field when $p$ is a prime number. Given a linear equation $ax + b = c$ where $a\neq 0,b,$ and $c$ are elements of your field, there exists a unique solution given by $x = (c - b)a^{-1},$ as for any nonzero element $a$ in a field there is a unique element $a^{-1}$ such that $a a^{-1} = a^{-1}a = 1.$