Question

Solve in positive integers:
$7x+4y+19z=84$.

Answer 1

$7x+4y+19z=84$

The maximum value of $z=\left[\frac{84}{19}\right]=4$

If $z=4,7x+4y=8$. This will not have any positive intergral values of $x$ and $y$

If $z=3,7x+4y=27\Rightarrow x=1$ and $y=5$.

if $z=2,7x+4y=46\Rightarrow x=2$ and $y=8$.

if $z=2,7x+4y=46\Rightarrow x=6$ and $y=1$.

if $z=1,7x+4y=65\Rightarrow x=3$ and $y=11$.

if $z=1,7x+4y=65\Rightarrow x=7$ and $y=4$