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When I type this equation ($6x + 25 = 7y$) into WolframAlpha, it is able to tell me that the integer solution for this equation is:

$x = 7n + 4$, $y = 6n + 7$, where n in the set of all integers

How can I arrive at this solution on my own?

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    $\begingroup$ Hint $\ (7 - 6 = 1)\ $ times $25.\ $ Generally use the [Extended Euclidean Algorithm](math.stackexchange.com/a/85841/242) to get the Bezout identity $\, ja + kb = c\,$ for $\,\gcd(a,,b) = c\,$ then scale that as need be. $\endgroup$ Commented Jan 29, 2019 at 20:34
  • $\begingroup$ Plug in the solution and see what the result is. Can you now do the backwards operation? $\endgroup$ Commented Jan 29, 2019 at 20:36
  • $\begingroup$ @PeterChikov what if I don't know the solution? $\endgroup$ Commented Jan 29, 2019 at 20:41
  • $\begingroup$ And see this answer for the form of the general solution (= particular + homogeneous solution) $\endgroup$ Commented Jan 29, 2019 at 20:45
  • $\begingroup$ Thanks @BillDubuque I will take a look at these! $\endgroup$ Commented Jan 29, 2019 at 20:48

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Since $6\mid 6y-24$ and $6\mid 7y-25$ we have $$6\mid (7y-25)-(6y-24)=y-1$$

Thus $y-1 = 6t$ for some integer $t$, so $\boxed{y= 6t+1}$ and pluging in $6x+25=7y$ we get: $$6x+25 =42t+7\implies \boxed{x= 7t-3}$$


It looks on first sight that I got different solution, but that is not true.

Puting $n=t-1$ we get $$ y=6t+1= 6(t-1)+7=6n+7$$ and $$x= 7(t-1)+4=7n+4$$

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  • $\begingroup$ What does the vertical pipe symbol mean? Is there an article I can read to learn about that? $\endgroup$ Commented Jan 30, 2019 at 11:07
  • $\begingroup$ $a\mid b$ means $a$ divide $b$, so $3\mid 12$ is true, while $8\mid 4$ is not. $\endgroup$
    – nonuser
    Commented Jan 30, 2019 at 11:08
  • $\begingroup$ The only potential problem I might see with this is that I don't think $7y - 25$ will always be divisible by $6$. For example, when $y = 0$. $\endgroup$ Commented Feb 1, 2019 at 11:09
  • $\begingroup$ Is 0 of form 6n+7? $\endgroup$
    – nonuser
    Commented Feb 1, 2019 at 13:09
  • $\begingroup$ Ahh I see now. My bad. Please disregard my last comment. $\endgroup$ Commented Feb 1, 2019 at 13:37

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