$x + 2y + 3z = 4$
$w = x + 2\times y$, then the equation becomes $w + 3z = 4$. $\gcd(1, 3) = 1 | 4$, so this two variable equation is solvable. $w = -2, y = 2$
i can't seem to pass this point
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What exactly confuses you? – 3x89g2 May 17 '15 at 19:31
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Have a look here, or here. For $ax+by+cz=d$ see here. – Dietrich Burde May 17 '15 at 19:31
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why does w = -2 n y= 2, n how'd i get the general solution for w and z? – user241340 May 17 '15 at 19:35
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Please see http://www.millersville.edu/~bikenaga/number-theory/linear-diophantine/linear-diophantine.html – 3x89g2 May 17 '15 at 19:42
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i've been looking at that example the past hour, i don't get the part when ((3, 5) = 1 | 10, so this two variable equation is solvable. w = 5, y = −1, is a particular solution, so the general solution is w = 5s + 5, z = −3s −1. – user241340 May 17 '15 at 19:44
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Are you convinced that if $\gcd(3,5)$ divides $10$ then the equation is solvable? – 3x89g2 May 17 '15 at 19:48
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yes, i am convinced – user241340 May 17 '15 at 19:50
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@user241340 So I guess you are not convinced that we have $\textit{all}$ the solutions? – 3x89g2 May 17 '15 at 19:54
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can you please explain then? – user241340 May 17 '15 at 19:56
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Suppose we have a Diophantine equation $ax+by=c$. If $\gcd(a,b)$ divides $c$, then this equation has solution.$w=-2, y=2$ is one. Then consider the Diophantine equation $x+2y=-2$. $\gcd(1,2)=1$, which divides $-2$ again. So you then solve this equation.
3x89g2
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