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I want to solve this equation: $3x+4y=14$

I present you what I have so far:

$\gcd(3,4)=1$ which is not $14$. I notice that: $3(6) + 4(-1) =14$

So using Bézout : $3(6-4k) + 4(-1+3k) = 14 (1)$, where $k$ integer. So we have $k>1/3$ and $k<3/2$. So $k = 1$. By replacing $k$ in equation $(1)$ we get: $a=2, b=2$, which indeed solves the equation. However, I dont get my previous solution $(a,b)=(4,-1)$ which is also correct.

Am I applying Bézout wrong? Or am I not supposed to find the solution that I used to find the new ones, if they exist? Do I have $2$ solutions and that's it or am I missing something? Thank you.

Bernard
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JamesTheProg
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    For a solution to exist, it is necessary that $\gcd(a,b)$ divides $d$. Then you can solve the case $ax + by = \gcd(a,b)$ and scale those solutions to get $d$ instead of $\gcd(a,b)$. – hardmath Jun 26 '21 at 13:59
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    Are $x$ and $y$ integers? – Bernard Jun 26 '21 at 14:22
  • There are infinitely many solutions. Take any solution $(x,y)$ of $3x+4y=1$; then $(14x,14y)$ is a solution of the given equation. – saulspatz Jun 26 '21 at 14:22
  • See the linked dupes - there are always infinitely many solutions when it is solvable. e.g $\bmod b!:\ ax\equiv d\Rightarrow a(x+nb)\equiv d\ \ $ – Bill Dubuque Jun 26 '21 at 21:44
  • "However, I dont get my previous solution (a,b)=(4,−1) which is also correct." Why should you get your previous solution? There are an infinite number of solutions so there is no reason to assume you will get the same solution every method. But you should get compatible solutions. – fleablood Jun 26 '21 at 23:52

1 Answers1

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$3x+4y=14$ its base solution is $x_0=2, y_0=2$. Take $x=2-4n,y=2+3n,$ where $n\in I$.

Z Ahmed
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