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This question is an exact duplicate of:

I am trying to find an $x$ and $y$ that solve the equation $15x - 16y = 10$, usually in this type of question I would use Euclidean Algorithm to find an $x$ and $y$ but it doesn't seem to work for this one. Computing the GCD just gives me $16 = 15 + 1$ and then $1 = 16 - 15$ which doesn't really help me. I can do this question with trial and error but was wondering if there was a method to it.

Thank you

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marked as duplicate by Bill Dubuque algebra-precalculus Nov 16 '18 at 22:51

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Can you solve the congruence $15x\equiv10\pmod{16}$? $\endgroup$ – Lord Shark the Unknown Nov 16 '18 at 22:17
  • $\begingroup$ $x$ is congruent to $6$ (mod $16$) ? but not sure how that helps me? thanks $\endgroup$ – ElMathMan Nov 16 '18 at 22:30
  • $\begingroup$ $x\equiv6\pmod{16}$ means $x=6+16k$ where $k\in\Bbb Z$. So then, what is $y$? $\endgroup$ – Lord Shark the Unknown Nov 17 '18 at 5:20
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Note that by Bezout's identity since $\gcd(15,16)=1$ we have

$$15\cdot (-1+k\cdot 16)+16 \cdot (1-k\cdot 15)=1 \quad k\in\mathbb{Z}$$

are all the solution for $15a+16b=1$ and from here just multiply by $10$.

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In this case you don't really need the full power of the Euclidean algorithm. Since you know $$ 16 - 15 = 1 $$ you can just multiply by $10$ to conclude that $$ 16 \times 10 + 15 \times(-10) = 10. $$ Now you have your $y$ and $x$.

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  • $\begingroup$ wouldn't this work for $16y$+$15x$ = $10$? $\endgroup$ – ElMathMan Nov 16 '18 at 22:33
  • $\begingroup$ It doesn't matter what you "call" $x$ and $y$. If you want $15x - 16y = 10$ just take $x = y = -10$ . $\endgroup$ – Ethan Bolker Nov 17 '18 at 0:20
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You have $16-15=1$

What about $$ x=-10+16k, y= -10+15k ?$$

That implies

$$ 15 x-16y=10$$ Which is a solution

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