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Determine all triples $(x,y,z)$ of integers satisfying the equation $3x+4y+5z=6$

I am not familiar with Diophantine equations with more variables. How do I solve this? Please anyone suggest some easier way to solve these linear Diophantine equations.
I think this can help here -
If $(x_0, y_0)$ are solutions to $ax + by = C$ with $gcd(a,b) = 1$, then $(x_0 + bt, y_0 - at)$ for all integer $t$ are also solutions.

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Note that we have that $$3x+4y=6-5z$$ So from what you know, we have that $(x,y)$ is characterized by $(-2+z-4t,3-2z+3t)$ for some integer $t$. So $(x,y,z)$ are for $t,z \in \mathbb{Z}$, $$(-2+z-4t,3-2z+3t,z)$$

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    $\begingroup$ Explain a little bit more... how to get the initial solution $(x,y)$ ? $\endgroup$ – Rezwan Arefin Feb 3 '17 at 6:52
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    $\begingroup$ @RezwanArefin Well, to find the initial solution we separate $x$ and $y$ into their $z$ parts, so $x=x_{0}+zx_{1}$, $y=y_{0}+zy_{1}$ that $3x_{0}+4y_{0}=6$ and $3x_{1}+4y_{1}=-5$. Using this, we can get the initial solution. $\endgroup$ – S.C.B. Feb 3 '17 at 7:11
  • $\begingroup$ With another approach I got $(1 + 3s + 4t, 1 -s-3t, 1-s), s,t \in \mathbb{Z}$. Is it right also? The approach uses the fact that $3x + 4y \equiv 1\; \text{mod}\; 5$ or $3x + 4y = 1+5s$ $\endgroup$ – Rezwan Arefin Feb 3 '17 at 8:11
  • $\begingroup$ @Rezwan Well, that does follow from my characterization, but Yes, that works as well. $\endgroup$ – S.C.B. Feb 3 '17 at 9:54

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