Show that the system of linear equations x+y+z=p, 2x−y+3z=q, 4x+y+5z=r is inconsistent if r != 2p+q.

The system of equations is:

1. $x+y+z=p$
2. $2x-y+3z=q$
3. $4x+y+5z=r$

Multiply equation (1) by 2 and add it to equation (2):

We have derived the equation $4x+y+5z = 2p+q$.
The third equation given in the system is $4x+y+5z = r$.

If the system is to have a solution, then the derived equation must be consistent with the third equation. This means that the right-hand sides must be equal:

However, the problem states that the system is inconsistent if $r \neq 2p+q$.
If $r \neq 2p+q$, then we have a contradiction:
$4x+y+5z = 2p+q$
$4x+y+5z = r$
Since $2p+q \neq r$, the same expression $4x+y+5z$ cannot simultaneously be equal to two different values, $2p+q$ and $r$. Therefore, no values of $x, y, z$ can satisfy all three equations simultaneously.

Thus, the system of equations is inconsistent if $r \neq 2p+q$.