You just find $y$

If one solves the above 2x3 linear system, the solution set is:

$(x,y,z) = (2+k, 241-2k, k); k \in \mathbb{N}.$

If we require $x > y > z > 78$ , then we have:

$2 + k > 241 - 2k \Rightarrow k > \frac{239}{3} \approx 79.67$

$241 - 2k > k \Rightarrow k < \frac{241}{3} \approx 80.33$

Hence $z = k = 80$ is the only positive integer solution that satisfies both inequalities. The desired solution is just $y = 241 - 2(80) = \boxed{81}.$

$x + y + z = 243 \Rightarrow z = 243 - x -y$

Substitute $z$ into the second equation

$\begin{aligned} 5x + 6y + 7(243-x-y) & = 1456 \\ 5x+ 6y+ 1701 -7x-7y & = 1456 \\ 2x + y & = 245 \\ x + \color{#D61F06} x+ y \color{#333333} & = 245 \quad \color{#3D99F6} \Rightarrow x+y = 243-z\\ x +243 - z & = 245 \\ x -z & = 2 \quad \Rightarrow \color{#20A900} x= z+2 \\ x+ y+ z & = 243 \\ 2z + y & = 241 \\ y & = 241-2z \end{aligned}$

Now, since $x>y>z>78$ , we can do something with those inequalities.

$\begin{aligned} x & > y \\ 2+ z & > 241- 2z \\ 3z & > 239 \quad \Rightarrow z > 79.67 \end{aligned}$

Another inequation,

$\begin{aligned} y & > z \\ 241-2z & > z \\ 3z & < 241 \quad \Rightarrow z<80.33 \end{aligned}$

Now, we have $79.67 <z< 80.33$ . Since $z$ is an integer, we have $z= 80$ , which leads us to find that $x= 82$ and $y= \boxed{81}$ .

Suppose that $A=x+y+z=243$ $B=5x+6y+7z=1456$ Then, we get $6A-B=6(x+y+z)-(5x+6y+7z)=6.243-1456$ $6A-B=x-z=1458-1456$ $6A-B=x-z=2$ $6A-B=x=2+z$ Because $x$ , $y$ , and $z$ are positive integers and $x>y>z$ . We can conclude that $x$ , $y$ , and $z$ are consecutive positive integers Where $y=z+1$ so $x+y+z=z+2+z+1+z=243$ $3z=240$ $z=80$ $z+1=81$ $y=81$