$ 4x -5y + 24z = 4A $, $ 2x - 2y + 2z = 10$. What is the largest possible value of $A$? If $x,y,z$ integers that satisfy
 $$ 4x -5y + 24z = 4A $$
$$ 2x - 2y + 2z = 10$$
with 
$y < 2x$ and $y-20z< 0$, what is the largest possible value of $A$?

Attempt:
We can rewrite the equations as:
$$ -y + 20z  = 4A - 20 $$
$$ 2x - 2y + 2z = 10$$
or
$$ -y + 20z = 4A - 20$$
$$ x - 19z = 25 - 4A $$
so we get $y - 20z = 20 - 4A < 0 $, so $A > 5$. Also, we can obtain
$$ -y + 20z  = 4A - 20 $$
$$ 2x - (19/10)y  = 12 - (2/5)A$$
so we have general solutions
$$ z = A/5 -1 + y/20$$
$$ x = 6 - A/5 + (19y/20) $$
how to continue find the maximum possible value of $A$?
 A: Solving the system
$$
\begin{cases}
4x-5y+24z=a
\qquad\qquad\;\;\;
\\[4pt]
2x-2y+2z=10\\
\end{cases}
$$
for $x,y$ yields
$$
\begin{cases}
x=19z+25-4a
\qquad\qquad\;\;\;
\\[4pt]
y=20z+20-4a\\
\end{cases}
\\[18pt]
$$
\begin{align*}
\text{Then}\;\;&y - 20z < 0\\[4pt]
\iff\;&(20z+20-4a)-(20z)  < 0\\[4pt]
\iff\;&20-4a < 0\\[4pt]
\iff\;&a > 5\\[4pt]
\iff\;&a \ge 6\\[10pt]
\text{and}\;\;&y < 2x\\[4pt]
\iff\;&20z+20-4a < 2(19z+25-4a)\\[4pt]
\iff\;&10z+10-2a < 19z+25-4a\\[4pt]
\iff\;&9z+15 > 2a\\[4pt]
\iff\;&z > \frac{2a-15}{9}\\[4pt]
\end{align*}
Thus, for any integer $a \ge 6$, and any integer $z > {\large{\frac{2a-15}{9}}}$, we can let
$$
\begin{cases}
x=19z+25-4a
\qquad\qquad\;\;\;
\\[4pt]
y=20z+20-4a\\
\end{cases}
$$
and all the specified conditions are satisfied.

Hence, while there is a minimum value for $a$, namely $a=6$, there is no maximum value for $a$.
A: Eliminate $y$ with $y=x+z-5$. The constraints become
$$-x+z<5,\\x-19z<5$$ and the objective function is
$$4A=4x-5y+25z=29x+20y-125.$$
As the constraints include an unbounded part of the first quadrant, $A$ is unbounded above !
(For instance, $x=n,y=2n-5,z=n$ satisfy the constraints and yield $4A=49n-125$.)
