Diophantine equation with three variables The question is:
Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of $6548 for their tickets on a particular flight, how many of each type of ticket
were sold?
Now I set up my equation as 
$140x+110y+78z=6548$
But I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for 
$x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n)$
Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations. 
 A: Combining the comments:
$140x+110y+78z=6548$
and
$x + y + z = 69$
$\implies 78x + 78y + 78z = 69*78 = 5382$
$\implies 62x + 32y  = 1166 \implies 31x + 16y = 69*78 = 583$
And we can quickly deduce that $x = 9, y = 19, z = 41$ (by simple inspection in my case - using that we only have integer values for $x,y,z$.
A: If the $\gcd$ of the ticket prices does not divide the total revenue, then you are correct that there will be no integer solution. However you are not immediately guaranteed a solution if the $\gcd$ does divide the revenue, because we are constrained to non-negative numbers of tickets. So we could potentially run into a Frobenius-coin-type failure.
Here the total number of tickets reduces this to a simple "two-coin" problem:
\begin{align}
&&140x+110y+78z &= 6548\\
\text{divide by }\gcd(x,y,z)=2&& 70x+55y+39z &= 3274\\
&&x+y+z &= 69\\
\text{multiply by }39 && 39x+39y+39z &= 2691\\
\text{subtract eqns} && 31x+16y &= 583\\
\bmod 16 && 31x\equiv 15x \equiv -1x&\equiv 583\equiv 7\\
\bmod 16 && x&\equiv -7\equiv 9\\
\text{test }x=25 && 31\cdot25 &= 775>583 \\
\text{thus }x=9 && 31\cdot 9 +16y&= 583 \\
 && y= (583-279)/16  &= 19\\
 && z= 69-(19+9)  &= 41\\
\end{align}
In the reduced equation $31x+16y = 583$, since $583>(31{-}1)\cdot (16{-}1)$ the coin problem issue could not apply - the total is big enough to guarantee a solution.
