Find the value of $c$ which makes it possible to solve:
$u+v+2w=2,$ $2u+3v-w=5,$ $3u+4v+w=c$
Find the value of $c$ which makes it possible to solve:
$u+v+2w=2,$ $2u+3v-w=5,$ $3u+4v+w=c$
HINT: Add the two first equations.
Set up your augmented matrix in the usual way:
$\left[\begin{array}{rrr|r} 1&1&2&2\\ 2&3&-1&5\\ 3&4&1&c \end{array}\right]\;.$
Then row-reduce it; reducing the first column, for instance, yields
$\left[\begin{array}{rrr|c} 1&1&2&2\\ 0&1&-5&1\\ 0&1&-5&c-6 \end{array}\right]\;.\tag{1}$
Now you can either stop and think about the equations corresponding to the bottom two rows of $(1)$ (what does $c$ have to be in order for them to be consistent?), or finish the row-reduction and then think about what $c$ has to be to avoid having an inconsistent system.
If you add the second and third equation you get 5u+7v=5+c ; if you multiply the second equation by 2 and then add it to the first then you get 4u+u+6v+v=12=5u+7v , so c=7