# How to solve the equation $15x- 16y= 10$ [duplicate]

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

## marked as duplicate by Bill Dubuque algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 16 '18 at 22:51

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

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

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 $$10$$.

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 $$x$$.

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

You have $$16-15=1$$

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

That implies

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