All values of $x$

Asked May 09 '23 at 11:01

Active May 09 '23 at 11:45

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So I just learned about modular arithmetic today and I was solving a system of congruences probably that you saw before. the statement was

$n ≡ 3\pmod5$

$n ≡ 1\pmod 7$

$n ≡ 6\pmod8$

writing this again as $$5x + 3 = 7y + 1 = 8z + 6$$ and trying to get a equality $8z + 6 = 7y + 1$

$8z + 5 = 7y$

so $1z + 5$ is divisible by $7$ which means all solutions for $z$ is $7n' + 2$ this will be the LHS or RHS of the equality

but this time I dont want to solve for z but I want to solve for x to get something more likeable than $3b ≡ 0\pmod5$ and I dont want to use the lcd function or find the first solution and add $8x'$ to it trick

$5x + 3 = 8z + 6$

$5x = 8z + 3$

all solution for $x$ ?

edited May 09 '23 at 11:45

asked May 09 '23 at 11:01

This is hard to follow. You appear to be using the same variable ($x$) to represent two different things. Please edit for clarity. And to explain why you don't want to use the standard methods to solve the problem. – lulu May 09 '23 at 11:04
2

This may help (or not). – TheSilverDoe May 09 '23 at 11:36
1

$5x = 8z + 3\Longrightarrow 3(z+1) = 5(x-z)\Longrightarrow z +1 \equiv 0 \mod 5$ – eyeballfrog May 09 '23 at 11:57

All values of $x$

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