Four equations are not enough to determine five unknowns, so it shouldn't be surprising that you can't solve this system. Just leave the last unknown arbitrary, and write the others in terms of it. For instance, if your unknowns are x,y,z,w,v, then your last row says that -9w+37v=-1, so that w=(37/9)v-1/9. Proceed as usual with your back substitution, except that instead of getting numbers for each unknown, you'll get expressions in v.
What this is telling you is that v could be anything at all, with the other values depending on what v is.