There are number of different ways to solve this: using Cramer's rule, Rouché–Capelli theorem, some forms of matrix decomposition... But probably the best way for you would be to simply thy to solve the system and see what you get.

`x+2y+hz=h`

`x+3y+4z=h`

`2x+4y+5z=5`

Now 1) subtract first row...

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There are number of different ways to solve this: using Cramer's rule, Rouché–Capelli theorem, some forms of matrix decomposition... But probably the best way for you would be to simply thy to solve the system and see what you get.

`x+2y+hz=h`

`x+3y+4z=h`

`2x+4y+5z=5`

Now 1) subtract first row from the second row and

2) subtract first row multiplied by 2 from the third row

`x+2y+hz=h`

`0+y+(4-h)z=0`

`0+0+(5-2h)z=5-2h`

Now, only way we would have some problem with solving this system of equations is if `5-2h=0=>h=5/2` (then the last row would be only zeros).

If we put that value of `h` back into our system we would get infinitely many solutions, to be exact solutions would be defined with

`y=15/2-3x`

`z=-5+2x,` `x in RR`

In all the other cases the system has unique solution. And since a system of linear equations is inconsistent when it has no solution **there exist no real number h for which this system is inconsistent**.